A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>In the diagram, $PQS$ lies on a circle and $PCBQ$ lies on another circle. $PQ$ is a common chord of the two circles. $BPS$ is a straight line. $QC$ is a tangent to the circle $PQS$ at $Q$. $QC$ and $BP$ intersect at $M$.</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/1782643685536-ow48s6vb88.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM2ODU1MzYtb3c0OHM2dmI4OC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzA1LCJleHAiOjE3OTI3MjYwMDV9.FVZSyQf0yiQfklkw-km8jPQXBZJHvrr6yCHgDbqOrGQ" style="max-width: min(320px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a-4" data-list-item-path="item-a-4"><div class="list-item-content"><span class="list-item-text">Prove that&nbsp;$BC$ is parallel to $QS$.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b-4" data-list-item-path="item-b-4"><div class="list-item-content"><span class="list-item-text">Prove that triangle $PQM$ is similar to triangle $QSM$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c-4" data-list-item-path="item-c-4"><div class="list-item-content"><span class="list-item-text">Hence show that&nbsp;$QM^2 = PM \times SM$.</span><span class="inline-marks">[2]</span></div></li></ol>

In the diagram, $PQS$ lies on a circle and $PCBQ$ lies on another circle. $PQ$ is a common chord of the two circles. $BPS$ is a straight line. $QC$ is a tangent to the circle $PQS$ at $Q$. $QC$ and $BP$ intersect at $M$.

Prove that triangle $PQM$ is similar to triangle $QSM$.

Alternate Segment Theorem

Angles in the Same Segment

Similarity

Angles Formed by Two Parallel Lines and a Transversal

2025 DSS PRELIMS P2 Q3

Proofs in Plane Geometry

<p>In the diagram, $TA$ and $TB$ are tangents from $T$ to the circle at $A$ and $B$, respectively. $C$ is a point on the circle and $D$ lies on $BC$ such that&nbsp;$CD = AD$. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="207" data-height="164" data-proportional style="width:207px;height:164px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:207px;--graph-aspect:207 / 164;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles $TAB$ and $DAC$ are similar,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle $ATB + \text{angle}\hspace{0.5em} ADB = 180^\circ$, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral $ADBT$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $A$, $D$, $B$ and $T$.</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">Hence, or otherwise, prove that&nbsp;$TD$ is parallel to $AC$.</div></li></ol>

In the diagram, $TA$ and $TB$ are tangents from $T$ to the circle at $A$ and $B$, respectively. $C$ is a point on the circle and $D$ lies on $BC$ such that $CD = AD$. Prove that

triangles $TAB$ and $DAC$ are similar,

angle $ATB + \text{angle } ADB = 180^\circ$, and

quadrilateral $ADBT$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $A$, $D$, $B$ and $T$.

Hence, or otherwise, prove that $TD$ is parallel to $AC$.

Properties of Special Triangles

Proving Concyclicity

Equal Tangents from an External Point

<p>In the diagram, $TA$ and $TB$ are tangents from $T$ to the circle with centre $O$ at $A$ and $B$, respectively. $AC$ is a diameter of the circle. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="198" data-height="163" data-proportional style="width:198px;height:163px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:198px;--graph-aspect:198 / 163;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle $AOB = 2 \times$ angle $ACB$,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$CB$ is parallel to $OT$,</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">triangles $OBC$ and $TAB$ are similar, and</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">quadrilateral $AOBT$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $A$, $O$, $B$ and $T$.</div></li></ol>

In the diagram, $TA$ and $TB$ are tangents from $T$ to the circle with centre $O$ at $A$ and $B$, respectively. $AC$ is a diameter of the circle. Prove that

triangles $OBC$ and $TAB$ are similar, and

quadrilateral $AOBT$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $A$, $O$, $B$ and $T$.

Angles at the Centre and Circumference

Tangents Perpendicular to Radii

<p>In the diagram, $O$ is the centre of the larger circle. $AOB$, $AKC$ and $BKD$ are straight lines. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="220" data-height="199" data-proportional style="width:220px;height:199px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:220px;--graph-aspect:220 / 199;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$DK = CK$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AB$ is parallel to $CD$ , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$DKB$ bisects angle $ODC$ .</div></li></ol>
<p>As shown, $A,B,C,D$ lie on the larger circle and $A,O,D,K$ lie on the smaller circle.</p>

In the diagram, $O$ is the centre of the larger circle. $AOB$, $AKC$ and $BKD$ are straight lines. Prove that

As shown, $A,B,C,D$ lie on the larger circle and $A,O,D,K$ lie on the smaller circle.

<p>In the diagram, $O$ is the centre of the smaller circle. $ABC$ and $AED$ are straight lines. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="252" data-height="198" data-proportional style="width:252px;height:198px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:252px;--graph-aspect:252 / 198;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$AC = AD$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$BE$ is parallel to $CD$.</div></li></ol>
<p>As shown in the diagram, $A,B,O,D$ lie on one circle, and $B,C,D,E$ lie on the circle with centre $O$.</p>

In the diagram, $O$ is the centre of the smaller circle. $ABC$ and $AED$ are straight lines. Prove that

As shown in the diagram, $A,B,O,D$ lie on one circle, and $B,C,D,E$ lie on the circle with centre $O$.

Angles in Cyclic Quadrilaterals

<p>In the diagram, the two circles intersect at $B$ and $E$. Given that&nbsp;$ABC$ and $FED$ are straight lines, prove that&nbsp;$AF$ is parallel to $CD$.</p>
<div class="graph-block" data-graph-config="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" data-width="237" data-height="193" data-proportional style="width:237px;height:193px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:237px;--graph-aspect:237 / 193;"></div>

In the diagram, the two circles intersect at $B$ and $E$. Given that $ABC$ and $FED$ are straight lines, prove that $AF$ is parallel to $CD$.

<p>In the diagram, $AP$ and $AQ$ are tangents to the smaller and larger circles at $A$. Given that&nbsp;$QBR$ is a straight line, prove that</p>
<div class="graph-block" data-graph-config="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" data-width="182" data-height="141" data-proportional style="width:182px;height:141px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:182px;--graph-aspect:182 / 141;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$RP$ is parallel to $AQ$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles $RPM$ and $QAM$ are similar.</div></li></ol>

In the diagram, $AP$ and $AQ$ are tangents to the smaller and larger circles at $A$. Given that $QBR$ is a straight line, prove that

triangles $RPM$ and $QAM$ are similar.

<p>In the diagram, $ABCD$ is a cyclic quadrilateral with $AD = BC$. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="178" data-height="166" data-proportional style="width:178px;height:166px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:178px;--graph-aspect:178 / 166;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$BD = AC$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">quadrilateral $ABCD$ is a trapezium.</div></li></ol>

In the diagram, $ABCD$ is a cyclic quadrilateral with $AD = BC$. Prove that

Properties of Special Quadrilaterals

Congruence

<p>In the figure, two circles intersect at $B$ and $E$. $ABC$ and $DEF$ are straight lines. Prove that&nbsp;$AF$ is parallel to $CD$.</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/1768111433813-wtjqcfh6i7i.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE0MzM4MTMtd3RqcWNmaDZpN2kucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.PylOVuLoOidk8ZbJxM_mrMxE2rywufn9DMgG98foRA8" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the figure, two circles intersect at $B$ and $E$. $ABC$ and $DEF$ are straight lines. Prove that $AF$ is parallel to $CD$.

<p>In the figure, $O$ is the centre of the circle $BCDE$ and $ABC$ and $AED$ are straight lines, prove that</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-1780398684393-mvhu24dmu" data-list-item-path="item-1780398684393-mvhu24dmu"><div class="list-item-paragraph">$AD=AC$,</div></li><li data-list-item-id="item-1780398684841-fkn4cdoq8" data-list-item-path="item-1780398684841-fkn4cdoq8"><div class="list-item-paragraph">Given that&nbsp;$AB=AE$, show that&nbsp;$BE$ is parallel to $CD$.</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/1768111313403-sdv4ekep96g.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTEzMTM0MDMtc2R2NGVrZXA5NmcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.MQF_zr95I1XWW8PUj21RAPBhW7-x6EAJdttgmyIwhjE" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>
<p>As shown in the diagram, $A,B,O,D$ lie on one circle, and $B,C,D,E$ lie on the circle with centre $O$.</p>

In the figure, $O$ is the centre of the circle $BCDE$ and $ABC$ and $AED$ are straight lines, prove that

<p>In the figure, $TA$ and $TB$ are tangents to the circle, centre $O$, from $T$.</p>
<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;"><ol class="i-list"><li><div class="list-item-paragraph">Given that&nbsp;$AOC$ is a diameter of the circle, state a reason why $\angle AOB=2\angle ACB$.</div></li></ol><p>Hence, prove that</p><ol class="i-list" start="2"><li><div class="list-item-paragraph">$CB$ is parallel to $OT$,</div></li><li><div class="list-item-paragraph">A circle with diameter $OT$ can be drawn to pass through $A$ and $B$,</div></li><li><div class="list-item-paragraph">$\Delta OBC$ is similar to $\Delta TAB$.</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/1768110672714-c5voub3uhv7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA2NzI3MTQtYzV2b3ViM3VodjcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.M_4dQgVXJpEblv2Lka5hUWaqPPvyIMxJRbfmscI6SJ8" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, $TA$ and $TB$ are tangents to the circle, centre $O$, from $T$.

<p>In the given diagram, $AP$ is tangent at $A$ to the smaller circle $AQB$, and $AQ$ is tangent at $A$ to the larger circle $APRB$ and $QBR$ is a straight line. Prove that&nbsp;$RP$ is parallel to $AQ$.</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/1768110489011-xeqgigjp67e.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA0ODkwMTEteGVxZ2lnanA2N2UucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.JzBsLUPxymCzhIaDNp8EuJyNnDyQgD724qtO_gBIikg" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the given diagram, $AP$ is tangent at $A$ to the smaller circle $AQB$, and $AQ$ is tangent at $A$ to the larger circle $APRB$ and $QBR$ is a straight line. Prove that $RP$ is parallel to $AQ$.

<p>In the diagram, $EFG$ is a straight line. The points $E$, $F$ and $G$ are the midpoints of $AD$, $AC$ and $BC$ respectively. Prove that</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">$EG = \frac{1}{2}(AB + DC)$,</div></li><li><div class="list-item-paragraph">$\angle CAB = \angle ACD$.</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/1768108652165-vzdx6az529k.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg2NTIxNjUtdnpkeDZhejUyOWsucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.KqdgvIkLi0QSjaoK1hduy3rg7KI9yUXdoFjb0hC8Jm0" style="max-width: min(180px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $EFG$ is a straight line. The points $E$, $F$ and $G$ are the midpoints of $AD$, $AC$ and $BC$ respectively. Prove that

Midpoint Theorem

<p>Given that&nbsp;$AB$ is parallel to $CD$. Find the value(s) of the unknown(s) in each of the figures below, stating your reasons clearly.<br><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="i-list"><li data-list-item-id="item-1787554920013-l5pidwo85" data-list-item-path="item-1787554920013-l5pidwo85"><div class="content-block list-item-image" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/06/Extra-Practice-Q7-i-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNi9FeHRyYS1QcmFjdGljZS1RNy1pLWRpYWdyYW0ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.iOCU2CAlX8GLq02wkJHR1_dNaXLlxFoK-3XhtSYO82A" alt="List item image" class="alignleft custom-width" style="max-width: min(200px, 100%); height: auto;" /></div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list" start="2"><li data-list-item-id="item-1787554931135-ra4upeenu" data-list-item-path="item-1787554931135-ra4upeenu"><div class="content-block list-item-image" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/migrated/2024/06/Extra-Practice-Q7-ii-diagram.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNi9FeHRyYS1QcmFjdGljZS1RNy1paS1kaWFncmFtLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI3MjU3MDUsImV4cCI6MTc5MjcyNjAwNX0.Tp9yun0I7xL-3X2-vjrZNKgbvkCg6KSUhPPN9B_RgRk" alt="List item image" class="aligncenter custom-width" style="max-width: min(150px, 100%); height: auto;" /></div></li></ol></div></div></div>

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



Basic Geometry

<p>In the figure below, $ABCD$ is a trapezium, where $AB$ is parallel to $CD$ and $AD = CD$. Produce $CB$ such that $AB = BE$. Given that angle $ADC = 94^\circ$, angle $ABC = 58^\circ$, answer the following questions:</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/1767685404303-v1no5pwdqsl.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2ODU0MDQzMDMtdjFubzVwd2Rxc2wucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.c8LcNrOlSWL1VXg350SclrmaLnVaU8MduRANcozNgQs" 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-paragraph">Show that $AC$ is the bisector of angle $BAD$.</div></li></ol>
<p>Find the sizes of</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">angle $AEB$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $DAE$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $BCD$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $ACB$.</span></div></li></ol>

In the figure below, $ABCD$ is a trapezium, where $AB$ is parallel to $CD$ and $AD = CD$. Produce $CB$ such that $AB = BE$. Given that angle $ADC = 94^\circ$, angle $ABC = 58^\circ$, answer the following questions:

 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2ODUyMjQ4MDktenV0MTAxOGE3bC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzA1LCJleHAiOjE3OTI3MjYwMDV9.2euR61vNJHEpqRwViLwSW2FYi4dkKJs6t-qMV7PH1hQ" 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

Triangle Angle Sum and Exterior Angles

Triangles and Quadrilaterals

<p>In the figure below, triangle $ABC$ is an isosceles triangle, where&nbsp;$AD=CD$ . Given that&nbsp;$AD$ is parallel to $BC$ , angle $BAC=76^{\circ}$ , angle $CDF=56^{\circ}$ , and $ADF$ , $CAE$ are straight lines, find the sizes of<br><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/1767683615445-06tossrzzkxo.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2ODM2MTU0NDUtMDZ0b3Nzcnp6a3hvLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI3MjU3MDUsImV4cCI6MTc5MjcyNjAwNX0.2py_tP7OjD-urZDjefS8V8GcXk1QOAnTNS8f3sAKaPM" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-1767683620908-r15i3layi" data-list-item-path="item-1767683620908-r15i3layi"><div class="list-item-paragraph">angle $DAC$,</div></li><li data-list-item-id="item-1767683621388-8d3a4298s" data-list-item-path="item-1767683621388-8d3a4298s"><div class="list-item-paragraph">angle $ABC$,</div></li><li data-list-item-id="item-1767683621956-etmm4t8ns" data-list-item-path="item-1767683621956-etmm4t8ns"><div class="list-item-paragraph">angle $EAB$,</div></li><li data-list-item-id="item-1767683622804-edo7jcio3" data-list-item-path="item-1767683622804-edo7jcio3"><div class="list-item-paragraph">angle $EAD$,</div></li></ol>

In the figure below, triangle $ABC$ is an isosceles triangle, where $AD=CD$ . Given that $AD$ is parallel to $BC$ , angle $BAC=76^{\circ}$ , angle $CDF=56^{\circ}$ , and $ADF$ , $CAE$ are straight lines, find the sizes of



<p>Given that&nbsp;$ABCD$ is a trapezium, where&nbsp;$AD$ is parallel to $BC$, find the value(s) of the unknown(s) in each of the following figures, stating your reasons clearly.</p>
<ol class="i-list"><li data-list-item-id="item-1767681260971-foun60kqx" data-list-item-path="item-1767681260971-foun60kqx"><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-7-i-diagram-2.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi03LWktZGlhZ3JhbS0yLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI3MjU3MDUsImV4cCI6MTc5MjcyNjAwNX0.-0ptgbIEr4mwKJh9-pOMC0sikvO1Z4-xx0j9gT8OSSI" alt="List item image" class="alignleft custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li><li data-list-item-id="item-1767681261538-cc76w3cte" data-list-item-path="item-1767681261538-cc76w3cte"><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-7-ii-diagram-1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi03LWlpLWRpYWdyYW0tMS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzA1LCJleHAiOjE3OTI3MjYwMDV9.tcH0jL6ntc_mRpAk2jnaeeDrEoIeHfngVF2pjD-6QDA" alt="List item image" class="alignleft custom-width" style="max-width: min(300px, 100%); height: auto;" /></div></li></ol>

Given that $ABCD$ is a trapezium, where $AD$ is parallel to $BC$, find the value(s) of the unknown(s) in each of the following figures, stating your reasons clearly.

Quadrilaterals

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0zLWEtZGlhZ3JhbS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzA1LCJleHAiOjE3OTI3MjYwMDV9.HCoUpktH2ggqh4OpiKeAtEhP4uSHlCsEk1BGNwNRzlo" 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvbWlncmF0ZWQvMjAyNC8wNS9RdWVzdGlvbi0zLWItZGlhZ3JhbS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzA1LCJleHAiOjE3OTI3MjYwMDV9.nNLyXGDp9Ttubo3Yiih9Av7Go-ddr76JuIimkdWNsD8" 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>In the figure below, $AB \perp PQ$, where $E$ is the foot of perpendicular; $CD \perp PQ$, where $F$ is the foot of perpendicular. <strong>Using this figure</strong>, show that $AB \parallel CD$, i.e., if two lines are perpendicular to a third line, then they are parallel to each other.</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/1767679333252-w9bsat12jzi.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2NzkzMzMyNTItdzlic2F0MTJqemkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTcwNSwiZXhwIjoxNzkyNzI2MDA1fQ.MKHCku3dA6m9jCn7SIt5jlPxxYzTUB_t16jA7_RB8QQ" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>

In the figure below, $AB \perp PQ$, where $E$ is the foot of perpendicular; $CD \perp PQ$, where $F$ is the foot of perpendicular. **Using this figure**, show that $AB \parallel CD$, i.e., if two lines are perpendicular to a third line, then they are parallel to each other.

Math Question Bank