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

<p>In the diagram, $TA$ is the tangent to the circle at $A$ and $BC$ is a diameter of the circle. $S$ is a point on $TA$ produced and $OS$ intersects $AB$ at $M$. The circle that passes through $A$, $M$ and $S$ has its centre at the midpoint of $AS$. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="288" data-height="193" data-proportional style="width:288px;height:193px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:288px;--graph-aspect:288 / 193;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles $ABT$ and $CAT$ are similar,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles $AMS$ and $CAB$ are similar, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral $ASOC$ is a trapezium.</div></li></ol>

In the diagram, $TA$ is the tangent to the circle at $A$ and $BC$ is a diameter of the circle. $S$ is a point on $TA$ produced and $OS$ intersects $AB$ at $M$. The circle that passes through $A$, $M$ and $S$ has its centre at the midpoint of $AS$. Prove that

triangles $ABT$ and $CAT$ are similar,

triangles $AMS$ and $CAB$ are similar, and

Alternate Segment Theorem

Properties of Special Quadrilaterals

Similarity

Angles in a Semicircle and the Converse

Proofs in Plane Geometry

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

Angles in Cyclic Quadrilaterals

Angles in the Same Segment

Congruence

Angles Formed by Two Parallel Lines and a Transversal

<p>In triangle $ABC$, $M$ and $N$ are the midpoints of $AB$ and $AC$, respectively. $PMNQ$ is a straight line such that&nbsp;$BP$, $CQ$ are perpendicular to $PMNQ$. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">quadrilateral $BCQP$ is a rectangle, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">area of triangle $ABC =$ area of quadrilateral $BCQP$.</div></li></ol>

In triangle $ABC$, $M$ and $N$ are the midpoints of $AB$ and $AC$, respectively. $PMNQ$ is a straight line such that $BP$, $CQ$ are perpendicular to $PMNQ$. Prove that

quadrilateral $BCQP$ is a rectangle, and

area of triangle $ABC =$ area of quadrilateral $BCQP$.

Midpoint Theorem

<p>In parallelogram $ABCD$, $E$, $F$ are on diagonal $AC$ such that&nbsp;$AE = CF$. $BEH$ and $DFG$ are straight lines. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles $BAE$ and $DCF$ are congruent,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles $BAH$ and $DCG$ are congruent, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral $BGDH$ is a parallelogram.</div></li></ol>

In parallelogram $ABCD$, $E$, $F$ are on diagonal $AC$ such that $AE = CF$. $BEH$ and $DFG$ are straight lines. Prove that

triangles $BAE$ and $DCF$ are congruent,

triangles $BAH$ and $DCG$ are congruent, and

quadrilateral $BGDH$ is a parallelogram.

<p>In rectangle $ABCD$, $P$, $S$ are on the side $AD$. $BD$ and $BP$ divide angle $ABC$ into $3$ equal parts, $CS$ bisects angle $BCD$ and $BD$, $CS$ intersect at $M$. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles $ABP$ and $CBD$ are similar,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AD = 3AP$,</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$AD = \left( 6 + 3\sqrt{3} \right)PS$, and</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">$CM = \sqrt{3}MS$.</div></li></ol>

In rectangle $ABCD$, $P$, $S$ are on the side $AD$. $BD$ and $BP$ divide angle $ABC$ into $3$ equal parts, $CS$ bisects angle $BCD$ and $BD$, $CS$ intersect at $M$. Prove that

triangles $ABP$ and $CBD$ are similar,

$AD = \left( 6 + 3\sqrt{3} \right)PS$, and

<p>In the rhombus $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $AD$, respectively. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">quadrilateral $PQRS$ is a rectangle.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Generalise the above result by stating one sufficient condition that quadrilateral $ABCD$ must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of $ABCD$ is a rectangle.</div></li></ol>

In the rhombus $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $AD$, respectively. Prove that

Generalise the above result by stating one sufficient condition that quadrilateral $ABCD$ must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of $ABCD$ is a rectangle.

<p>In rectangle $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $AD$, respectively. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">quadrilateral $PQRS$ is a rhombus.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Generalise the above result by stating one sufficient condition that quadrilateral $ABCD$ must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of $ABCD$ is a rhombus.</div></li></ol>

In rectangle $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $AD$, respectively. Prove that

Generalise the above result by stating one sufficient condition that quadrilateral $ABCD$ must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of $ABCD$ is a rhombus.

<p>In quadrilateral $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $AD$, respectively. Prove that quadrilateral $PQRS$ is a parallelogram.</p>
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In quadrilateral $ABCD$, $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $BC$, $CD$ and $AD$, respectively. Prove that quadrilateral $PQRS$ is a parallelogram.

<p>In quadrilateral $ABCD$, $P$, $Q$, $R$ and $S$ are on $AD$, $BD$, $BC$ and $AC$, respectively, such that quadrilateral $PQRS$ is a parallelogram. Given that&nbsp;$AP : PD = BQ : QD = 3 : 1$, prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$AS : SC = BR : RC = 3 : 1$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$PS = QR = \frac{3}{4}DC$.</div></li></ol>

In quadrilateral $ABCD$, $P$, $Q$, $R$ and $S$ are on $AD$, $BD$, $BC$ and $AC$, respectively, such that quadrilateral $PQRS$ is a parallelogram. Given that $AP : PD = BQ : QD = 3 : 1$, prove that

<p>In triangle $ABC$, $BQ$ and $CP$ intersect at $O$. $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $AC$, $OC$ and $OB$ respectively. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$PQRS$ is a parallelogram,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$OS = \frac{1}{3} BQ$, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$BS = \frac{1}{3} BQ$.</div></li></ol>

In triangle $ABC$, $BQ$ and $CP$ intersect at $O$. $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $AC$, $OC$ and $OB$ respectively. Prove that

<p>In the diagram, $BFD$, $AFC$ and $BEC$ are straight lines. It is given that&nbsp;$AF = FC$, $BF = FD$ and $BE = EC$. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="257" data-height="139" data-proportional style="width:257px;height:139px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:257px;--graph-aspect:257 / 139;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">$FE = \frac{1}{2} AB$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AB = DC$.</div></li></ol>

In the diagram, $BFD$, $AFC$ and $BEC$ are straight lines. It is given that $AF = FC$, $BF = FD$ and $BE = EC$. Prove that

<p>In the diagram, $E$, $F$, $G$ and $H$ are the midpoints of the sides of the rectangle $ABCD$. The points $P$, $Q$, $R$ and $S$ are the midpoints of $HE$, $EF$, $FG$ and $GH$ respectively. By drawing the line segments $EG$ and $HF$, 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">$EF = EH$,</div></li><li><div class="list-item-paragraph">$EFGH$ is a rhombus,</div></li><li><div class="list-item-paragraph">$\angle SPQ = 90{}^\circ$,</div></li><li><div class="list-item-paragraph">$PQRS$ is a rectangle.</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/1768109085978-hw9eg7wq788.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDkwODU5NzgtaHc5ZWc3d3E3ODgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ.6xCNYBRezg7ZEpDulfSz2zx73uyZnhhzeKp4PPIWRlM" style="max-width: min(240px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $E$, $F$, $G$ and $H$ are the midpoints of the sides of the rectangle $ABCD$. The points $P$, $Q$, $R$ and $S$ are the midpoints of $HE$, $EF$, $FG$ and $GH$ respectively. By drawing the line segments $EG$ and $HF$, prove that

Applications of Pythagoras’ Theorem

<p>Quadrilateral $QUAD$ has an inscribed quadrilateral $PROF$, where&nbsp;$P$, $R$, $O$ and $F$ are the midpoints of $QD$, $QU$, $UA$ and $AD$ respectively. Prove that quadrilateral $PROF$ is a parallelogram.</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/1768109018332-puqlk12l57f.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDkwMTgzMzItcHVxbGsxMmw1N2YucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ._hri_mYEN2KioJo1xjbF-m2z71FvZZJb8FlMZAnhH94" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

Quadrilateral $QUAD$ has an inscribed quadrilateral $PROF$, where $P$, $R$, $O$ and $F$ are the midpoints of $QD$, $QU$, $UA$ and $AD$ respectively. Prove that quadrilateral $PROF$ is a parallelogram.

<p>In $\Delta ABC$, $BQ$ and $CP$ intersect at $O$. $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $AC$, $OC$ and $OB$ respectively. Prove that&nbsp;$PQRS$ is a parallelogram. Hence, show that&nbsp;$OS = \frac{1}{3} BQ$.</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/1768108924862-6o0dkhqg89l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg5MjQ4NjItNm8wZGtocWc4OWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ.OVF7zQgqWLUXzGgHnJh2_1lxMIabNon8nxm9Yonu6S4" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In $\Delta ABC$, $BQ$ and $CP$ intersect at $O$. $P$, $Q$, $R$ and $S$ are the midpoints of $AB$, $AC$, $OC$ and $OB$ respectively. Prove that $PQRS$ is a parallelogram. Hence, show that $OS = \frac{1}{3} BQ$.

Proving Quadrilaterals

<p>In the diagram, $AD = DC$, $AE = EB$, and the lines $BD$ and $CE$ of $\Delta ABC$ intersect at $G$. The points $F$ and $H$ are the midpoints of $BG$ and $CG$ 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">$DEFH$ is a parallelogram,</div></li><li><div class="list-item-paragraph">$BF = DG$ and $CH = GE$.</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/1768108734383-v2cbjs5694.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg3MzQzODMtdjJjYmpzNTY5NC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjY3LCJleHAiOjE3OTI2NjI5Njd9.VeCuZiWPVXw1ERqJDZIvb0RF4g_GzlUxDqkhxf4N8vA" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, $AD = DC$, $AE = EB$, and the lines $BD$ and $CE$ of $\Delta ABC$ intersect at $G$. The points $F$ and $H$ are the midpoints of $BG$ and $CG$ respectively. Prove that

<p>In the diagram, $ABCD$ is a rectangle and $PC$ is perpendicular to $BD$. 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><div class="list-item-paragraph">$\Delta ABD$ is similar to $\Delta PDC$,</div></li><li><div class="list-item-paragraph">$AB^2 = BD \times PD$.</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/1768107673191-kz2bpzpuyzl.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDc2NzMxOTEta3oyYnB6cHV5emwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ.EBe560q48JPVyDil6E9krhemywg8cjo-_lys8au9ggI" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $ABCD$ is a rectangle and $PC$ is perpendicular to $BD$. Prove that

<p>In the figure below, <strong>ABCDE</strong> is a regular pentagon, <strong>ABMN</strong> and <strong>AEQP</strong> are squares.</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/1767769077295-ftyyzy06ow4.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc3NjkwNzcyOTUtZnR5eXp5MDZvdzQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ.EGdODkBIcZEeMtW5xkxd4YcBgHgLmV6oDc4lCqF9D7k" style="max-width: min(350px, 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">Find the size of angle <strong>BAE</strong>.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the size of angle <strong>CDF</strong>.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the size of angle <strong>BFE</strong>.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the size of angle <strong>NAP</strong>.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the size of angle <strong>ABF</strong>.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the size of angle <strong>QEF</strong>.</span></div></li><li><div class="list-item-content"><span class="list-item-text">Hence determine the type of quadrilateral <strong>ABFE</strong>, stating your reasons clearly.</span></div></li></ol>

In the figure below, **ABCDE** is a regular pentagon, **ABMN** and **AEQP** are squares.

Hence determine the type of quadrilateral **ABFE**, stating your reasons clearly.

Angles in Regular Polygons

Polygons

<p>In the figure below, $ABCD$ and $ACEF$ are rhombuses, where $AE$ intersects $CD$ and $CF$ at $M$ and $N$, respectively. Given that angle $ACB = 26^\circ$, 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/1767768855941-woh8nihn5r.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc3Njg4NTU5NDEtd29oOG5paG41ci5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNjYyNjY3LCJleHAiOjE3OTI2NjI5Njd9.eiYph-H-LP6Uqah-T7E1FH1eGisOogVva_0CeFbiL_M" 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 $CEN$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $CDF$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $DMN$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $MCN$.</span></div></li><li><div class="list-item-paragraph">Suppose now, given that angle $ACB \neq 26^\circ$ and triangle $CMN$ is a right-angled isosceles triangle, where $CN = MN$.</div><div class="list-item-paragraph">Find the new size of angle $ACB$.</div></li></ol>

In the figure below, $ABCD$ and $ACEF$ are rhombuses, where $AE$ intersects $CD$ and $CF$ at $M$ and $N$, respectively. Given that angle $ACB = 26^\circ$, find the sizes of:

Triangles, Quadrilaterals

<p>In the figure below, $ABCD$ is a rhombus. Produce $CB$ to $E$ such that $BD = BE$. Given that angle $BCD = 60°$, 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/1767768565074-t29rr082xul.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc3Njg1NjUwNzQtdDI5cnIwODJ4dWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ.ceWM-vibjoMV7c2tCphGiJcsdad6JsQ0UEb3gw7MTvc" style="max-width: min(300px, 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 $ADB$,</span></div></li><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 $BED$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $BFE$.</span></div></li><li><div class="list-item-paragraph">Determine the type of triangles $ABD$ and $BCD$.</div></li><li><div class="list-item-paragraph">Construct the line $AE$. Determine the type of quadrilateral $AEBD$, stating your reasons clearly.</div></li></ol>

In the figure below, $ABCD$ is a rhombus. Produce $CB$ to $E$ such that $BD = BE$. Given that angle $BCD = 60°$, find the sizes of

<p>In the figure below, $ABCD$ is a kite, where the diagonals $AC, BD$ intersect at $O$. $E$ is a point on $BD$, such that $AECD$ is a rhombus. Given that angle $EAO = 33^\circ$, angle $BCF = 87^\circ$, where $F$ is a point on the extension of $DC$, 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/1767685675672-jf2j3h56sdi.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2ODU2NzU2NzItamYyajNoNTZzZGkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY2NywiZXhwIjoxNzkyNjYyOTY3fQ.QeqQVjIb0NFSbPF2-0TjYC0zXLpkoSs0u0Cpzie8-K4" style="max-width: min(300px, 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 $ADO$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $AEC$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $DCE$,</span></div></li><li><div class="list-item-content"><span class="list-item-text">angle $BCE$.</span></div></li><li><div class="list-item-paragraph">Hence determine the type of triangle $ABC$, stating your reasons clearly.</div></li></ol>

In the figure below, $ABCD$ is a kite, where the diagonals $AC, BD$ intersect at $O$. $E$ is a point on $BD$, such that $AECD$ is a rhombus. Given that angle $EAO = 33^\circ$, angle $BCF = 87^\circ$, where $F$ is a point on the extension of $DC$, find the sizes of

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