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PSLE

PSLE Foundation Mathematics

<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

<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/1781422877742-krek3774bzi.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE0MjI4Nzc3NDIta3JlazM3NzRiemkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU2MiwiZXhwIjoxNzkyNzI1ODYyfQ.WB5YGdIxYpaou_BiFacHP30vu_EKNHFHxzMGjXh3tY4" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The diagram shows a circle, centre $O$. Points $A$, $B$, $D$ and $E$ lie on the circle. <br>The line $AD$ is a diameter of the circle and $DCP$ is the tangent of the circle at $D$. <br>The lines $EB$ and $AD$ intersect at point $Q$. <br>The chord $AB$ is produced to meet the tangent $DCP$ at $P$. <br>It is given that&nbsp;$\text{angle}\hspace{0.5em} CBP = \text{angle}\hspace{0.5em} DEB$.</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Show that&nbsp;$\text{angle}\hspace{0.5em} BCD = 90^\circ$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Show that&nbsp;$DP^2 = BP^2 + DA \times BC$.<span class="inline-marks">[4]</span></div></li></ol>

The diagram shows a circle, centre $O$. Points $A$, $B$, $D$ and $E$ lie on the circle. 
The line $AD$ is a diameter of the circle and $DCP$ is the tangent of the circle at $D$. 
The lines $EB$ and $AD$ intersect at point $Q$. 
The chord $AB$ is produced to meet the tangent $DCP$ at $P$. 
It is given that $\text{angle } CBP = \text{angle } DEB$.

Applications of Pythagoras’ Theorem

Similarity

Angles in a Semicircle and the Converse

2024 CGSS P1 Q13

<p>In the diagram, $AB$ is a diameter of the circle with centre $O$. $DE$ and $BF$ are tangents to the circle at $C$ and $B$ respectively. $DCE$ and $BEF$ are straight lines. Prove that</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/1782643825419-7qc4gmhh4c5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM4MjU0MTktN3FjNGdtaGg0YzUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU2MiwiZXhwIjoxNzkyNzI1ODYyfQ.al9Imk2grGQrDYOuc7vALV7eCTavlfbetQxsuPbe4Qg" style="max-width: min(180px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-content"><span class="list-item-text">$\Delta ABC$ and $\Delta BFC$ are similar,</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-content"><span class="list-item-text">hence or otherwise, show that&nbsp;$\Delta CFE$ is an isosceles triangle.</span><span class="inline-marks">[4]</span></div></li></ol>

In the diagram, $AB$ is a diameter of the circle with centre $O$. $DE$ and $BF$ are tangents to the circle at $C$ and $B$ respectively. $DCE$ and $BEF$ are straight lines. Prove that

$\triangle ABC$ and $\triangle BFC$ are similar,

hence or otherwise, show that $\triangle CEF$ is an isosceles triangle.

Tangents Perpendicular to Radii

Equal Tangents from an External Point

2025 AHS PRELIMS P1 Q13

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM2ODU1MzYtb3c0OHM2dmI4OC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NTYyLCJleHAiOjE3OTI3MjU4NjJ9.Uz6__peaeTrlddaMJlA94KoqAsocsOu33vQxiDXjpYM" 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 $QSM$.

Angles Formed by Two Parallel Lines and a Transversal

2025 DSS PRELIMS P2 Q3

<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>Two circles touch each other at $A$. The chord $MN$ of the larger circle cuts the smaller circle at $P$ and $Q$. Prove that&nbsp;$\angle NAQ = \angle MAP$.</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/1770726548885-zot8xkegjz.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjY1NDg4ODUtem90OHhrZWdqei5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NTYyLCJleHAiOjE3OTI3MjU4NjJ9.bkHPnrmkNijz_CY35N3g_AOMG0a0H7JbfZtulXwvzEs" style="max-width: min(150px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

Two circles touch each other at $A$. The chord $MN$ of the larger circle cuts the smaller circle at $P$ and $Q$. Prove that $\angle NAQ = \angle MAP$.

IP2U14 Exercise Q4

<p>In the diagram, the two circles touch each other externally at $F$ and $CFT$ is the common tangent. Given that&nbsp;$ABC$, $CDE$ and $AFE$ 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">angle $CBF + \text{angle}\hspace{0.5em} CDF = 180^\circ$, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle $BCD + \text{angle}\hspace{0.5em} BFD = 180^\circ$.</div></li><li data-list-item-id="item-1770117831518-ug1rdrv6i" data-list-item-path="item-1770117831518-ug1rdrv6i"><div class="list-item-paragraph">Hence prove that there exists a circle that passes through $B$, $C$, $D$ and $F$.</div></li></ol>
<p>Based on the answers above, or otherwise, prove that</p>
<ol class="i-list" start="4"><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">triangles $BCD$ and $ECA$ are similar, and</div></li><li data-list-item-id="item-v" data-list-item-path="item-v"><div class="list-item-paragraph">there exists a circle that passes through $A$, $B$, $D$ and $E$.</div></li></ol>

In the diagram, the two circles touch each other externally at $F$ and $CFT$ is the common tangent. Given that $ABC$, $CDE$ and $AFE$ are straight lines, prove that

angle $CBF + \text{angle } CDF = 180^\circ$, and

angle $BCD + \text{angle } BFD = 180^\circ$.

Based on the answers above, or otherwise, prove that

triangles $BCD$ and $ECA$ are similar, and

there exists a circle that passes through $A$, $B$, $D$ and $E$.

Angles in Cyclic Quadrilaterals

Proving Concyclicity

<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>
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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 $EDT = 2 \times \text{angle}\hspace{0.5em} BAC$ ,</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 $B$.

Triangle Angle Sum and Exterior Angles

<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 $AC$.

<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>
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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 $AOB = 2 \times$ 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

<p>In the diagram, $BC$ is a diameter of the circle. Triangle $BCR$ is an isosceles triangle with $BC = BR$ and $CR$ intersects the circle at $A$. The tangent to the circle at $A$ intersects $BR$ at $S$. 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 $BSA = 90^\circ$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles $ABC$ and $SBA$ are similar, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">the circle that passes through $A$, $B$ and $S$ has its centre at the midpoint of $AB$.</div></li></ol>

In the diagram, $BC$ is a diameter of the circle. Triangle $BCR$ is an isosceles triangle with $BC = BR$ and $CR$ intersects the circle at $A$. The tangent to the circle at $A$ intersects $BR$ at $S$. Prove that

triangles $ABC$ and $SBA$ are similar, and

the circle that passes through $A$, $B$ and $S$ has its centre at the midpoint of $AB$.

<p>In the diagram, $SAT$ is the tangent to the circle at $A$. $ADB$ and $AEC$ are straight lines such that&nbsp;$B$ and $C$ are on the circle and $DE$ is parallel to $SAT$. 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 $ABC$ and $AED$ are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">quadrilateral $BDEC$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $B$, $D$, $E$ and $C$.</div></li></ol>

In the diagram, $SAT$ is the tangent to the circle at $A$. $ADB$ and $AEC$ are straight lines such that $B$ and $C$ are on the circle and $DE$ is parallel to $SAT$. Prove that

triangles $ABC$ and $AED$ are similar, and

quadrilateral $BDEC$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $B$, $D$, $E$ and $C$.

<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>
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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 $ABT$ 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

Properties of Special Quadrilaterals

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

<p>In the diagram, $TCS$ is the tangent to the circle at $C$. $TBA$ and $RCD$ are straight lines that are parallel to each other. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="280" data-height="237" data-proportional style="width:280px;height:237px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:280px;--graph-aspect:280 / 237;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle $BCT =$ angle $ACD$,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles $BTC$ and $DAC$ are similar, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$AD = BC$.</div></li></ol>

In the diagram, $TCS$ is the tangent to the circle at $C$. $TBA$ and $RCD$ are straight lines that are parallel to each other. Prove that

triangles $BTC$ and $DAC$ are similar, and

Congruence

<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. $AB$ is a diameter and is parallel to $CD$. $TDS$ and $TBR$ are tangents to the circle at $D$ and $B$, respectively. Given that&nbsp;$TDS$ intersects $OA$ produced at $S$, 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 $ADS =$ angle $TBC$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle $CBD = 90^\circ - 2 \times$ angle $SDA$ , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">triangles $SAD$ and $BCD$ are similar.</div></li></ol>

In the diagram, $ABCD$ is a cyclic quadrilateral. $AB$ is a diameter and is parallel to $CD$. $TDS$ and $TBR$ are tangents to the circle at $D$ and $B$, respectively. Given that $TDS$ intersects $OA$ produced at $S$, prove that

angle $CBD = 90^\circ - 2 \times$ angle $SDA$ , and

triangles $SAD$ and $BCD$ are similar.

<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>
<div class="graph-block" data-graph-config="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" data-width="233" data-height="148" data-proportional style="width:233px;height:148px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:233px;--graph-aspect:233 / 148;"></div>
<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.

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