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N(A) A Math

N(A) E Math

PSLE

<p>Points $A$, $B$, $C$ and $D$ lie on a circle. The tangent to the circle at $A$ meets $BD$ produced at $E$. $AE$ is parallel to $BC$.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<p>Prove that</p>
<ol class="i-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">$AB = AC$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">$\angle CDE = 2\angle ABC$.<span class="inline-marks">[3]</span></div></li></ol>

Points $A$, $B$, $C$ and $D$ lie on a circle. The tangent to the circle at $A$ meets $BD$ produced at $E$. $AE$ is parallel to $BC$.

Alternate Segment Theorem

Properties of Special Triangles

Angles in the Same Segment

2024 FMSS PRELIMS P1 Q3

Proofs in Plane Geometry

<p>$P$, $Q$ and $R$ are points that lie on a circle. The tangent to the circle at $R$ meets the line $PQ$ produced at $S$. Point $X$ is on the line $PQ$ such that&nbsp;$RX$ bisects $\angle PRQ$. Prove that&nbsp;$RS=XS$.</p>

$P$, $Q$ and $R$ are points that lie on a circle. The tangent to the circle at $R$ meets the line $PQ$ produced at $S$. Point $X$ is on the line $PQ$ such that $RX$ bisects $\angle PRQ$. Prove that $RS=XS$.

IP2U14 HOT Q2

Properties of Circle

Polygons

<p>$A$, $D$, $E$ and $F$ are points on the circle such that&nbsp;$AD = ED$. $ADB$, $AEC$ and $DFC$ are straight lines and $CD$ bisects $\angle BCE$.</p>
<p>$\angle BED = 34^\circ$ and $\angle BFC = 124^\circ$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Prove that</div><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">$AB$ is perpendicular to $CD$,</div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">$BCED$ is a cyclic quadrilateral.</div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Describe clearly another circle that will pass through $A$, $C$ and $D$.</div></li></ol>
<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/1770727124566-2340bxfgz36.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjcxMjQ1NjYtMjM0MGJ4Zmd6MzYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY3NiwiZXhwIjoxNzkyNjYyOTc2fQ.hgkHaKcRHC_RJ48oGdNqv9mayEjg4Xcg-VbC5lUPEwI" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>
<p>As shown in the diagram, $B,E,F$ are collinear, with $F$ between $B$ and $E$.</p>

$A$, $D$, $E$ and $F$ are points on the circle such that $AD = ED$. $ADB$, $AEC$ and $DFC$ are straight lines and $CD$ bisects $\angle BCE$.

$\angle BED = 34^\circ$ and $\angle BFC = 124^\circ$.

Describe clearly another circle that will pass through $A$, $C$ and $D$.

As shown in the diagram, $B,E,F$ are collinear, with $F$ between $B$ and $E$.

Angles in Cyclic Quadrilaterals

Angles in a Semicircle and the Converse

Proving Concyclicity

IP2U14 HOT Q1

<p>Chords $AC$ and $BD$, perpendicular to each other, intersect at $E$.</p>
<p>In $\Delta ADE$, the altitude from $E$ meets $AD$ at $F$. $FE$ produced meets $BC$ at $G$. Prove that&nbsp;$G$ is the midpoint of $BC$.</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/1770726144466-1u5dr35xwkc.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjYxNDQ0NjYtMXU1ZHIzNXh3a2MucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY3NiwiZXhwIjoxNzkyNjYyOTc2fQ.xwbjxP7z_mis22xsFWcKYJmd0QAldtnIDvBt8_srsvw" style="max-width: min(170px, 100%); height: auto; display: block;" alt="Geometric construction of perpendicular chords and an altitude within a circle" class="alignright custom-width" /></figure></div>

Chords $AC$ and $BD$, perpendicular to each other, intersect at $E$.

In $\Delta ADE$, the altitude from $E$ meets $AD$ at $F$. $FE$ produced meets $BC$ at $G$. Prove that $G$ is the midpoint of $BC$.

IP2U14 Exercise Q3

<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>
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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 $TAB$ 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 $AC$.

Similarity

Angles Formed by Two Parallel Lines and a Transversal

Equal Tangents from an External Point

<p>In the diagram, $O$ is the centre of the larger circle. $AOB$, $AKC$ and $BKD$ 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">$DK = CK$ ,</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.

Angles at the Centre and Circumference

<p>In the diagram, $O$ is the centre of the smaller circle. $ABC$ and $AED$ 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">$AC = AD$, 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 $O$.

<p>In the diagram, $TAR$ is the tangent to the circle at $A$. $DCA$ and $DOB$ are straight lines. It is given that angle $BAR = 3 \times \text{angle}\hspace{0.5em} OCB$, 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">angle $ODC = 2 \times \text{angle}\hspace{0.5em} OBC$,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle $COD$ is an isosceles triangle, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">angle $DCO = \text{angle}\hspace{0.5em} CAT + 90^\circ$.</div></li></ol>

In the diagram, $TAR$ is the tangent to the circle at $A$. $DCA$ and $DOB$ are straight lines. It is given that angle $BAR = 3 \times \text{angle } OCB$, prove that

angle $ODC = 2 \times \text{angle } OBC$,

triangle $COD$ is an isosceles triangle, and

angle $DCO = \text{angle } CAT + 90^\circ$.

Tangents Perpendicular to Radii

<p>In the diagram, the smaller circle has centre $O$ and intersects the larger circle at $B$ and $C$. The point $O$ also lies on the larger circle. $ABD$ and $ACE$ are straight lines, where&nbsp;$A$ is on the smaller circle and $D, E$ are on the larger circle. $F$ is the point of intersection of $BE$ and $CD$. 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">angle $AEB = 180^\circ - 2 \times$ angle $BAC$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle $ABE$ is an isosceles triangle, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">angle $DFE = 360^\circ - 3 \times$ angle $BAC$ .</div></li></ol>

In the diagram, the smaller circle has centre $O$ and intersects the larger circle at $B$ and $C$. The point $O$ also lies on the larger circle. $ABD$ and $ACE$ are straight lines, where $A$ is on the smaller circle and $D, E$ are on the larger circle. $F$ is the point of intersection of $BE$ and $CD$. Prove that

angle $AEB = 180^\circ - 2 \times$ angle $BAC$ ,

triangle $ABE$ is an isosceles triangle, and

angle $DFE = 360^\circ - 3 \times$ angle $BAC$ .

<p>In the diagram, $ABCD$ is a cyclic quadrilateral with $BC = CD$ and $BC$ is parallel to $AD$. The tangent to the circle at $B$ intersects $DA$ produced at $T$. 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">angle $ABT = \text{angle}\hspace{0.5em} BDC$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle $BAT$ is an isosceles triangle.</div></li></ol>

In the diagram, $ABCD$ is a cyclic quadrilateral with $BC = CD$ and $BC$ is parallel to $AD$. The tangent to the circle at $B$ intersects $DA$ produced at $T$. Prove that

triangle $BAT$ is an isosceles triangle.

<p>In the diagram, $TA$ is the tangent to the circle at $A$ and $TB$ is a straight line that intersects the circle at $B$ and $D$. $C$ is a point on $TB$ such that&nbsp;$TA = TC$. 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">$TA \times AB = TB \times AD$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AC$ bisects angle $BAD$.</div></li></ol>

In the diagram, $TA$ is the tangent to the circle at $A$ and $TB$ is a straight line that intersects the circle at $B$ and $D$. $C$ is a point on $TB$ such that $TA = TC$. Prove that

<p>In the diagram, $TA$ is the tangent to the circle at $A$ and $TB$ is a straight line that intersects the circle at $B$ and $D$. $C$ is a point on $TB$ such that&nbsp;$AC$ bisects angle $BAD$. 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 $TAD$ and $TBA$ are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle $TAC$ is an isosceles triangle.</div></li></ol>

In the diagram, $TA$ is the tangent to the circle at $A$ and $TB$ is a straight line that intersects the circle at $B$ and $D$. $C$ is a point on $TB$ such that $AC$ bisects angle $BAD$. Prove that

triangles $TAD$ and $TBA$ are similar, and

triangle $TAC$ is an isosceles triangle.

<p>In the diagram, $TA$ is the tangent to the circle at $A$. $TDE$ and $TCB$ are straight lines such that&nbsp;$AD = AE$. 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">angle $TDC = $ angle $TEA$ , and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$TDE$ bisects angle $ATB$ .</div></li></ol>

In the diagram, $TA$ is the tangent to the circle at $A$. $TDE$ and $TCB$ are straight lines such that $AD = AE$. Prove that

<p>In the diagram, $SA$ is the tangent to the circle with centre $O$ at $A$ and $AB$ is a diameter. $C$ is a point on the circle and the tangent to the circle at $C$ intersects $SA$ at $T$. $BCS$ is a straight line. 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">angle $CTS = 2 \times$ angle $CAT$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$ST = CT$ , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$SC \times OC = AC \times AT$ .</div></li></ol>

In the diagram, $SA$ is the tangent to the circle with centre $O$ at $A$ and $AB$ is a diameter. $C$ is a point on the circle and the tangent to the circle at $C$ intersects $SA$ at $T$. $BCS$ is a straight line. Prove that

<p>In the diagram, $TA$ and $TB$ are tangents to the circle from $T$ at $A$ and $B$, respectively. Given that&nbsp;$TA$ is parallel to $BC$, 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">triangle $ABC$ is an isosceles triangle, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AB^2 = TB \times CB$.</div></li></ol>

In the diagram, $TA$ and $TB$ are tangents to the circle from $T$ at $A$ and $B$, respectively. Given that $TA$ is parallel to $BC$, prove that

triangle $ABC$ is an isosceles triangle, and

<p>In the diagram, $ABC$ and $EDC$ are straight lines such that&nbsp;$B$ and $D$ are the midpoints of $AC$ and $EC$, 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">triangles $CBD$ and $CAE$ are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$CA = CE$.</div></li></ol>
<p>As shown in the diagram, $A,B,D,E$ lie on a circle.</p>

In the diagram, $ABC$ and $EDC$ are straight lines such that $B$ and $D$ are the midpoints of $AC$ and $EC$, respectively. Prove that

triangles $CBD$ and $CAE$ are similar, and

As shown in the diagram, $A,B,D,E$ lie on a circle.

<p>In the diagram, the two circles touch each other internally at $C$ and $CD$ is the common tangent. $AD$ and $BD$ are tangents to the larger and smaller circles, 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">triangles $BCD$ and $ACD$ are isosceles triangles, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle $ABD$ is an isosceles triangle.</div></li></ol>

In the diagram, the two circles touch each other internally at $C$ and $CD$ is the common tangent. $AD$ and $BD$ are tangents to the larger and smaller circles, respectively. Prove that

triangles $BCD$ and $ACD$ are isosceles triangles, and

triangle $ABD$ is an isosceles triangle.

<p>In the diagram, $BCE$ and $ADE$ are straight lines. Given that angle $BAD =$ angle $ACD$, 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">$BD = AD$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$CD$ bisects angle $ACE$.</div></li></ol>

In the diagram, $BCE$ and $ADE$ are straight lines. Given that angle $BAD =$ angle $ACD$, prove that

<p>In the diagram, $ABF$, $CAG$ and $EFG$ are straight lines. Given that&nbsp;$AG = AC = CF$ and $AF = FB$, 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 $AFG$ and $FBC$ are congruent, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle $BFE$ is an isosceles triangle.</div></li></ol>

In the diagram, $ABF$, $CAG$ and $EFG$ are straight lines. Given that $AG = AC = CF$ and $AF = FB$, prove that

triangles $AFG$ and $FBC$ are congruent, and

triangle $BFE$ is an isosceles triangle.

Congruence

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTEzMTM0MDMtc2R2NGVrZXA5NmcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjY2MjY3NiwiZXhwIjoxNzkyNjYyOTc2fQ.zgHqW0IUBAHpVir1pXgLZ8nu7X_rmz6Y20HX5yCr0IA" 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

Math Question Bank