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PSLE

<p>In the diagram, $TA$ and $TC$ are tangents from $T$ to the circle at $A$ and $C$, respectively. $B$ is a point on the circle and the tangent to the circle at $B$ intersects $TC$ and $TA$ at $D$ and $E$, respectively. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="233" data-height="181" data-proportional style="width:233px;height:181px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:233px;--graph-aspect:233 / 181;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle $EDT = 2 \times \text{angle}\hspace{0.5em} BAC$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle $ABC = 90^\circ + \frac{1}{2} \times \text{angle}\hspace{0.5em} DTE$ , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral $ACDB$ is NOT a cyclic quadrilateral, i.e., there exists no such a circle that passes through $A$, $C$, $D$ and $B$.</div></li></ol>

In the diagram, $TA$ and $TC$ are tangents from $T$ to the circle at $A$ and $C$, respectively. $B$ is a point on the circle and the tangent to the circle at $B$ intersects $TC$ and $TA$ at $D$ and $E$, respectively. Prove that

angle $EDT = 2 \times \text{angle } BAC$ ,

angle $ABC = 90^\circ + \frac{1}{2} \times \text{angle } DTE$ , and

quadrilateral $ACDB$ is NOT a cyclic quadrilateral, i.e., there exists no such a circle that passes through $A$, $C$, $D$ and $B$.

Alternate Segment Theorem

Triangle Angle Sum and Exterior Angles

Proving Concyclicity

Proofs in Plane Geometry

<p>In the diagram, $PT$ is a tangent to the semicircle $APB$. If&nbsp;$PB$ bisects $\angle ABT$, prove that&nbsp;$\angle BTP=90{}^\circ $.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768110913020-lbddcwd9orn.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA5MTMwMjAtbGJkZGN3ZDlvcm4ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.KAA1NYifiEXtet96gr4nyl9gF6E-fJAI54A8th7y60A" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram, $PT$ is a tangent to the semicircle $APB$. If $PB$ bisects $\angle ABT$, prove that $\angle BTP=90{}^\circ $.

Angles in a Semicircle and the Converse

<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>In the diagram, $\angle BAC = \angle ADB = 90{}^\circ$. Prove that</p><ol class="i-list"><li><div class="list-item-paragraph">$\angle ABD = \angle CAD$,</div></li><li><div class="list-item-paragraph">$\Delta ABD$ is similar to $\Delta CAD$,</div></li><li><div class="list-item-paragraph">$AD^2 = BD \times CD$.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768107598448-puwiopowd3.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDc1OTg0NDgtcHV3aW9wb3dkMy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.gueZrxh_LHonXiE_GAQv5gbhnscNgfN2JUkwuX4Txx0" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

Similarity

<p>In the figure below, $ABC$ is a triangle enclosed by straight lines $MN$, $PQ$ and $RS$, where angles such as angle $CAN$, angle $ABR$ and angle $ACS$ are known as exterior angles of triangle $ABC$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767768078907-mxh93o674t.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc3NjgwNzg5MDctbXhoOTNvNjc0dC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.IMhq5st0_BI8XP9miNkb3cHyl7b0P9dBnfyo6Z190YA" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Using this figure, show that angle $CAN = $ angle $ABC + $ angle $ACB$, i.e., the size of an exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle (remote interior angles), which is known as the Exterior Angle Theorem.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Apply the Exterior Angle Theorem on the other two exterior angles, angle $ABR$ and angle $ACS$, by stating the relationship between the exterior angles and corresponding interior angles.</span></div></li></ol>

In the figure below, $ABC$ is a triangle enclosed by straight lines $MN$, $PQ$ and $RS$, where angles such as angle $CAN$, angle $ABR$ and angle $ACS$ are known as exterior angles of triangle $ABC$.

Using this figure, show that angle $CAN = $ angle $ABC + $ angle $ACB$, i.e., the size of an exterior angle at a vertex of a triangle equals the sum of the sizes of the interior angles at the other two vertices of the triangle (remote interior angles), which is known as the Exterior Angle Theorem.

Apply the Exterior Angle Theorem on the other two exterior angles, angle $ABR$ and angle $ACS$, by stating the relationship between the exterior angles and corresponding interior angles.

Triangles

Polygons

<p>In the figure below, $ABCD$ is a right-angled trapezium, where $AD$ is parallel to $BC$ and $AB$ is perpendicular to $BC$. $E$ and $F$ are on sides $AD$ and $CD$, respectively. Given that angle $AEB = 40^\circ$, angle $CBF = 16^\circ$ and angle $BCF = 54^\circ$, find the sizes of</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767685224809-zut1018a7l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2ODUyMjQ4MDktenV0MTAxOGE3bC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.YVzXJ_CUMEILuTAwE_QVMxngb_SOKjFVEe1aplNDI0s" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">angle $ABC$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $EDF$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $EBF$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $BFD$.</span></div></li></ol>

In the figure below, $ABCD$ is a right-angled trapezium, where $AD$ is parallel to $BC$ and $AB$ is perpendicular to $BC$. $E$ and $F$ are on sides $AD$ and $CD$, respectively. Given that angle $AEB = 40^\circ$, angle $CBF = 16^\circ$ and angle $BCF = 54^\circ$, find the sizes of

Angles Formed by Two Parallel Lines and a Transversal

Triangles and Quadrilaterals

Basic Geometry

<p>Given that&nbsp;$AB$ is parallel to $CD$, find the value(s) of the unknown(s) in the following figures, stating your reasons clearly.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">The figure shows a triangle $ABC$ and a parallel line $CD$. Angle $\angle BAC = y^\circ$ and $\angle ABC = x^\circ$. Outside the triangle, an exterior angle at $C$ is $112^\circ$ and $\angle BCD = 38^\circ$. Find the values of $x$ and $y$.</div><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-3-a-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0zLWEtZGlhZ3JhbS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.dxL02seghzJAgfU0tS9Z2JpTaR2KBiv41MBY88kpJ0I" alt="List item image" class="alignleft custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The figure shows parallel lines $CD$ and $AB$ intersected by two transversals that meet at point $E$. Angle $\angle AEB = 100^\circ$, $\angle CEA = z^\circ$, and $\angle DEB = y^\circ$. Inside the triangle $ABE$, angle $\angle EAB = x^\circ$, and an exterior angle at $B$ is $143^\circ$. Find the values of $x$, $y$, and $z$.</div><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-3-b-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0zLWItZGlhZ3JhbS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.yZtEqbRwOZBQDeZikTs5JLEEE4QO1YaKBg3q0euYpOE" alt="List item image" class="alignleft custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li></ol>

Given that $AB$ is parallel to $CD$, find the value(s) of the unknown(s) in the following figures, stating your reasons clearly.

 Triangles

<p>Given that&nbsp;$AOB$ and $COD$ are straight lines, find the value(s) of the unknown(s) in the following figures, stating your reasons clearly.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">The figure shows two intersecting straight lines forming two triangles, $\triangle AOC$ and $\triangle BOD$. In $\triangle AOC$, $\angle ACO = 62^\circ$ and $\angle AOC$ is adjacent to an exterior angle of $143^\circ$. In $\triangle BOD$, $\angle BDO = 105^\circ$. Find the values of $x$ and $y$.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-2-a-diagram-1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0yLWEtZGlhZ3JhbS0xLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI2NjI2MjAsImV4cCI6MTc5MjY2MjkyMH0.hvAkX_uXNa7NGTkHqXPB7AodFWNEPCFltxlDl3L1Mls" alt="List item image" class="aligncenter custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">The figure shows two intersecting lines $AOB$ and $COD$. $\triangle AOD$ has an interior angle $\angle OAD = 28^\circ$ and an exterior angle of $y^\circ$ at $D$. $\triangle COB$ is an isosceles triangle where&nbsp;$CO = OB$, with $\angle OCB = x^\circ$. The angle $\angle DOB$ is given as&nbsp;$78^\circ$. Find the values of $x$ and $y$.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-2-b-diagram-1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0yLWItZGlhZ3JhbS0xLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI2NjI2MjAsImV4cCI6MTc5MjY2MjkyMH0.I-RGrdW_eL27uET4dBqgLoiOfmPUc7qDyY0dxb1u2uk" alt="List item image" class="aligncenter custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li></ol>

Given that $AOB$ and $COD$ are straight lines, find the value(s) of the unknown(s) in the following figures, stating your reasons clearly.

Properties of Special Triangles

<p>Find the value(s) of the unknown(s) in each of the following figures, stating your reasons clearly.</p>
<ol class="i-list"><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-1-i-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0xLWktZGlhZ3JhbS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.GW4K-JI_1S4sSjK-v7UJvXVO88mV5E2Yy3biOvxF0Lo" alt="List item image" class="alignleft custom-width" style="max-width: min(200px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-1-ii-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0xLWlpLWRpYWdyYW0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.GEv8pyjhO1fAaDY9rav1jcF_qtK0QXQkeB82Xjdf204" alt="List item image" class="alignleft custom-width" style="max-width: min(250px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-1-iii-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0xLWlpaS1kaWFncmFtLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI2NjI2MjAsImV4cCI6MTc5MjY2MjkyMH0.Jdv7Q59-GOW6zXJjJzrFV8OUZBzIPnyN38qKiJqNsOM" alt="List item image" class="alignleft custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-1-iv-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0xLWl2LWRpYWdyYW0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ._0xEVubrm_mT3xJWDzIw0yASitdykntdScASYr-7s9A" alt="List item image" class="alignleft custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-1-v-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0xLXYtZGlhZ3JhbS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9._wgRYzk5lQhQKYnJHAP8CMtDogyu2c0XhsnHEjutwVE" alt="List item image" class="alignleft custom-width" style="max-width: min(280px, 100%); height: auto;" /></div></li><li><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-1-vi-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0xLXZpLWRpYWdyYW0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.M7cSjNe2u_FQLcTVpPOYskTFZ5G1xCZjs5AUPR_IPRw" alt="List item image" class="alignleft custom-width" style="max-width: min(250px, 100%); height: auto;" /></div></li></ol>

Find the value(s) of the unknown(s) in each of the following figures, stating your reasons clearly.

<p>In the figure below, $PQ$ is parallel to $RS$, where $A$ is a point on $PQ$ and $B, C$ are two points on $RS$. Using this figure, show that the sum of the sizes of all the interior angles in any triangle is $180^\circ$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767679056236-pr7zcm6psp.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2NzkwNTYyMzYtcHI3emNtNnBzcC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.0DiRhnx6boPrxJHxVlttaLpJedH9Y7pPQ7cO0_ulpRg" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>

In the figure below, $PQ$ is parallel to $RS$, where $A$ is a point on $PQ$ and $B, C$ are two points on $RS$. Using this figure, show that the sum of the sizes of all the interior angles in any triangle is $180^\circ$.

 Angles Formed by Two Parallel Lines and a Transversal

<ol class="i-list"><li><div class="list-item-paragraph">In the figure, triangle $OMN$ is a right-angled isosceles triangle, where $OM$ is perpendicular to $NM$ and $OM = NM$. Given that $OQ$ is parallel to $MN$ and $\angle POQ = \frac{1}{2} \angle QOM$, show that $P$, $O$, and $N$ lie on the same straight line.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-23-i-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0yMy1pLWRpYWdyYW0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.5MeO2uaQQxyhZQtbqMw3BjvmjWjq7HPDlqwqRITFlzQ" alt="List item image" class="aligncenter custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li><li><div class="list-item-paragraph">In the figure, triangle $OMN$ is a right-angled isosceles triangle, where $OM$ is perpendicular to $NM$ and $OM = NM$. Given that $P$ lies on $NO$ produced and $\angle POQ = \frac{1}{2} \angle QOM$, show that $OQ$ is parallel to $MN$.</div><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/05/Question-23-ii-diagram-1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0yMy1paS1kaWFncmFtLTEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.2h8ocU1On_UCXbton13SQvR4A6V0172ysDGK-iMUeA8" alt="List item image" class="aligncenter custom-width" style="max-width: min(400px, 100%); height: auto;" /></div></li></ol>

In the figure, triangle $OMN$ is a right-angled isosceles triangle, where $OM$ is perpendicular to $NM$ and $OM = NM$. Given that $OQ$ is parallel to $MN$ and $\angle POQ = \frac{1}{2} \angle QOM$, show that $P$, $O$, and $N$ lie on the same straight line.

In the figure, triangle $OMN$ is a right-angled isosceles triangle, where $OM$ is perpendicular to $NM$ and $OM = NM$. Given that $P$ lies on $NO$ produced and $\angle POQ = \frac{1}{2} \angle QOM$, show that $OQ$ is parallel to $MN$.

<p>In the figure below, $DE$ is parallel to $FG$ and $AC$ is parallel to $EF$. The triangle $ABC$ is a right-angled triangle, where&nbsp;$AB$ is perpendicular to $BC$. Given that angle $BAC = 24^\circ$, angle $CDE = 66^\circ$ and angle $EFG = 137^\circ$, find the sizes of</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list"><li data-list-item-id="item-1787555166718-euubj0xkc" data-list-item-path="item-1787555166718-euubj0xkc"><div class="list-item-paragraph"><p>angle $ACB$,</p></div></li><li data-list-item-id="item-1787555167483-xeygdvhvx" data-list-item-path="item-1787555167483-xeygdvhvx"><div class="list-item-paragraph"><p>reflex angle $DEF$,</p></div></li><li data-list-item-id="item-1787555167986-pv1up1l3t" data-list-item-path="item-1787555167986-pv1up1l3t"><div class="list-item-paragraph"><p>angle $ACD$.</p></div></li><li data-list-item-id="item-1787555168720-py55d4pwm" data-list-item-path="item-1787555168720-py55d4pwm"><div class="list-item-paragraph"><p>Hence prove or falsify $AB$ is parallel to $CD$.</p></div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1767678528819-93fgupfz0q9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2Nzg1Mjg4MTktOTNmZ3VwZnowcTkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.r70gF6MR0N1ipV4xirBd0w0KtP9_6WVChbypqrosEkk" style="max-width: min(350px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the figure below, $DE$ is parallel to $FG$ and $AC$ is parallel to $EF$. The triangle $ABC$ is a right-angled triangle, where $AB$ is perpendicular to $BC$. Given that angle $BAC = 24^\circ$, angle $CDE = 66^\circ$ and angle $EFG = 137^\circ$, find the sizes of

<p>In the diagram below, $AE//CD$, $\angle BDC=25{}^\circ $ and $\angle BCD=80{}^\circ $.  The angles are shown in the diagram.  Stating your reasons clearly, find the</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">value of $x$,</div></li><li><div class="list-item-paragraph">value of $y$.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764308503897-7y08z2ligh9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQzMDg1MDM4OTctN3kwOHoybGlnaDkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.SgfEBfTynldDIzM8dZ_e-2QnwyPeM_kxGd4fF3ZC9vs" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram below, $AE//CD$, $\angle BDC=25{}^\circ $ and $\angle BCD=80{}^\circ $.  The angles are shown in the diagram.  Stating your reasons clearly, find the

Angles in a Trapezium with Parallel Sides  $𝐴𝐸 ∥ 𝐶𝐷$

<p>In the diagram below, $RST$ is a straight line and is parallel to $PQ$. $\angle PSQ=84{}^\circ $ and $\angle QST=65{}^\circ $.  Calculate, stating your reasons clearly,<br></p>
<ol class="abc-list"><li><div class="list-item-paragraph">$\angle a$.</div></li><li><div class="list-item-paragraph">$\angle b$.</div></li><li><div class="list-item-paragraph">$\angle c$.</div></li></ol>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764074585733-sy6op58rozp.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzQ1ODU3MzMtc3k2b3A1OHJvenAucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.6joGXDQDUIRUsWx3xvR-trvBslvEy6XuZVs_xOsEiLo" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>

In the diagram below, $RST$ is a straight line and is parallel to $PQ$. $\angle PSQ=84{}^\circ $ and $\angle QST=65{}^\circ $.  Calculate, stating your reasons clearly,


Angles in Parallel Lines with Transversals

<p>In the diagram, $BCD$ is a straight line and $AB$ is parallel to $CE$. $\angle CAB=66{}^\circ $ and $\angle ACD=130{}^\circ $.  Stating your reasons clearly, calculate the value of<br></p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li data-list-item-id="item-1777114881051-q3k0qky18"><div class="list-item-paragraph">$x$,</div></li><li data-list-item-id="item-1777114881656-tyeu2thst"><div class="list-item-paragraph">$y$.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764074348218-hf2uci4c3fp.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzQzNDgyMTgtaGYydWNpNGMzZnAucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.PZC6BF-Uz791SjMXXJLJFTvvWgR8OJUS4fK0-gkcY_E" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $BCD$ is a straight line and $AB$ is parallel to $CE$. $\angle CAB=66{}^\circ $ and $\angle ACD=130{}^\circ $.  Stating your reasons clearly, calculate the value of


Angles Formed by Parallel Lines and Intersecting Segments

<p>In the diagram below, $AB$ is parallel to $CD$.  Given the angles shown,</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><p>find</p><ol class="abc-list"><li><div class="list-item-paragraph">$x$,</div></li><li><div class="list-item-paragraph">$y$.</div></li></ol><p>stating your reasons clearly. </p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764073261813-jsc8kquy4pm.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzMyNjE4MTMtanNjOGtxdXk0cG0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.e4SmMjLYr0fIzkJ0rhqCpcBagW42dhTnn2nlxVVjVhU" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram below, $AB$ is parallel to $CD$.  Given the angles shown,

Finding Unknown Angles Using Parallel Line Properties

<p>Showing all workings and stating reasons clearly, find the value of the unknown in the diagram below.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764072783419-f42yiw6ck08.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzI3ODM0MTktZjQyeWl3NmNrMDgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.qzvuhYSOYPtlvegoCvF1Qi2S0izKsdeaKv2g4Uvi2w4" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

Showing all workings and stating reasons clearly, find the value of the unknown in the diagram below.

Angle Relationships in Intersecting Triangles

<p>$KMN$ is a triangle and $LMN$ is a straight line. $\angle NKM=60{}^\circ $ and $\angle KML=140{}^\circ $.  Find the value of $z$, stating your reasons clearly. <br></p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764072543478-ux3sy4ochx.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzI1NDM0NzgtdXgzc3k0b2NoeC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjIwLCJleHAiOjE3OTI2NjI5MjB9.dWGUeK2yNa6fKM_rVj7wd45mEwVfT7AWbihQ0OKT-lk" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>

$KMN$ is a triangle and $LMN$ is a straight line. $\angle NKM=60{}^\circ $ and $\angle KML=140{}^\circ $.  Find the value of $z$, stating your reasons clearly. 


Exterior Angle of a Triangle Problem

<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><p>In the diagram, $CDE$ and $ADF$ are straight lines. $AB$ is parallel to $CD$. $\angle BDC=40{}^\circ $ and $\angle ADB=80{}^\circ $.<br>Find</p><ol class="abc-list"><li data-list-item-id="item-1776758573969-qzargpyr2"><div class="list-item-paragraph">angle $BAD$,</div></li><li data-list-item-id="item-1776758574476-vn5yq1a37"><div class="list-item-paragraph">angle $EDF$,</div></li><li data-list-item-id="item-1776758575116-o32piqovy"><div class="list-item-paragraph">reflex angle $CDF$,</div></li></ol><p>stating your reasons clearly. </p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764072334799-mdmeixxnogr.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzIzMzQ3OTktbWRtZWl4eG5vZ3IucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.2JvikEeYHcXdg0jDhxf5rKdKhjirrFXvG2aqeBEIFA4" style="max-width: min(300px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

Angle Relationships with Parallel Lines and Transversals

<p>In the diagram, $ABC$ is a triangle.  By stating the reasons clearly in your working, find the value of $x$.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764071373988-arunqn1zmi4.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNzEzNzM5ODgtYXJ1bnFuMXptaTQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.JeMSVYEhZynWFKqIcA4AbeaKzWZoKgIqNXVhz56mXAY" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram, $ABC$ is a triangle.  By stating the reasons clearly in your working, find the value of $x$.

Angle Sum of a Triangle

<p>In the diagram given below, $PQ$ and $TU$ are parallel and $WYM$ and $NXM$ are straight lines. Stating your reasons clearly, determine the values of $a$, $b$ and $c$.</p>
<div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1764049399886-17ar7qptpdt.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjQwNDkzOTk4ODYtMTdhcjdxcHRwZHQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjYyMCwiZXhwIjoxNzkyNjYyOTIwfQ.3Su3WARaeEbSpeWEhrMwLvgrupqtJ6J2nnuw59f_0Og" style="max-width: min(400px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram given below, $PQ$ and $TU$ are parallel and $WYM$ and $NXM$ are straight lines. Stating your reasons clearly, determine the values of $a$, $b$ and $c$.

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