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<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 $BC$.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE0MjI4Nzc3NDIta3JlazM3NzRiemkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.1Flp_NsMlZs4mwn5UlmjvHtAaEG9z5n1_EyDsWyH-fs" 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 $\text{angle } CBP = \text{angle } DEB$.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM4MjU0MTktN3FjNGdtaGg0YzUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.1z2ZUG6WZ5gu9FYEFDg4VrseSZvXCMVGD7Y33jQk7hw" 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.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM2ODU1MzYtb3c0OHM2dmI4OC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjkzNDYxLCJleHAiOjE3ODk2OTM3NjF9.YEBUfCEmozDaCxujae8aPytsVzDerm4BRqnvm5AUX-E" 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 $QSM$.

2025 DSS PRELIMS P2 Q3

<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>
<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/1770117776995-9jmd3gxsx3h.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTc3NzY5OTUtOWptZDNneHN4M2gucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.yRXqe9eFf-ViGIOZ5COLbMDu8X20ucqMf6lj3nRgfjA" style="max-width: min(270px, 100%); height: auto; display: block;" alt="Geometry problem showing two externally tangent circles and intersecting lines." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $CBF + \text{angle}\hspace{0.5em} CDF = 180^\circ$, and</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">angle $BCD + \text{angle}\hspace{0.5em} BFD = 180^\circ$.</span></div></li><li data-list-item-id="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"><div class="list-item-content"><span class="list-item-text">triangles $BCD$ and $ECA$ are similar, and</span></div></li><li data-list-item-id="item-v"><div class="list-item-content"><span class="list-item-text">there exists a circle that passes through $A$, $B$, $D$ and $E$.</span></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 $E$.

<p>In the diagram, $TA$ and $TC$ are tangents from $T$ to the circle at $A$ and $C$, respectively. $B$ is a point on the circle and the tangent to the circle at $B$ intersects $TC$ and $TA$ at $D$ and $E$, respectively. Prove that</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770117642202-3o7u7b9g2aq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTc2NDIyMDItM283dTdiOWcyYXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.UH-YAmuLH_iNFAs6wnbcrFklWRkvzAXF3OWtTYGrmlQ" style="max-width: min(220px, 100%); height: auto; display: block;" alt="Geometric diagram showing a circle with tangents TA, TC, and a third tangent DE intersecting TA and TC." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $EDT = 2 \times \text{angle}\hspace{0.5em} BAC$ ,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">angle $ABC = 90^\circ + \frac{1}{2} \times \text{angle}\hspace{0.5em} DTE$ , and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">quadrilateral $ACDB$ is NOT a cyclic quadrilateral, i.e., there exists no such a circle that passes through $A$, $C$, $D$ and $B$.</span></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 $B$.

<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="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770117499805-vctyje4xko.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTc0OTk4MDUtdmN0eWplNHhrby5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjkzNDYxLCJleHAiOjE3ODk2OTM3NjF9.YC0wqSta8-tnlxVx5akf7xRFsd4Ab3b02D95U4JQOFg" style="max-width: min(190px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">triangles $TAB$ and $DAC$ are similar,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">angle $ATB + \text{angle}\hspace{0.5em} ADB = 180^\circ$, and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">quadrilateral $ADBT$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $A$, $D$, $B$ and $T$.</span></div></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">Hence, or otherwise, prove that&nbsp;$TD$ is parallel to $AC$.</span></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>
<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/1770117338592-2woosw3eip7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTczMzg1OTItMndvb3N3M2VpcDcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.CdBVpVy82PGk2ZhGb5i1dLzE4pclpQi_8VhPxRNKxkw" style="max-width: min(180px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $AOB = 2 \times$ angle $ACB$,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">$CB$ is parallel to $OT$,</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">triangles $OBC$ and $TAB$ are similar, and</span></div></li><li data-list-item-id="item-iv"><div class="list-item-content"><span class="list-item-text">quadrilateral $AOBT$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $A$, $O$, $B$ and $T$.</span></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 $T$.

<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>
<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/1770117132490-rp0f3h1yy9q.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTcxMzI0OTAtcnAwZjNoMXl5OXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.JUmOrDsLNqyeuo6nUxfKV5Xspab7m_v6RKCduSuGzuY" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $BSA = 90^\circ$ ,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangles $ABC$ and $SBA$ are similar, and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">the circle that passes through $A$, $B$ and $S$ has its centre at the midpoint of $AB$.</span></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>
<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/1770116967799-7390b4h62o2.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTY5Njc3OTktNzM5MGI0aDYybzIucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.xPITHB1zgWfwt-nx1DGDEXMX_LMZoAx_4-JxhU4xZA4" style="max-width: min(200px, 100%); height: auto; display: block;" alt="Geometric diagram showing a circle with tangent SAT and parallel line DE" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">triangles $ABC$ and $AED$ are similar, and</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">quadrilateral $BDEC$ is a cyclic quadrilateral, i.e., there exists a circle that passes through $B$, $D$, $E$ and $C$.</span></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, $O$ is the centre of the larger circle. $AOB$, $AKC$ and $BKD$ 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/1770116831867-b9ushn2ih85.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTY4MzE4NjctYjl1c2huMmloODUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.-xCWSflzskLF4iNl8akd5dfYUtO-3VC9WUMyTv-y3SY" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">$DK = CK$ ,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">$AB$ is parallel to $CD$ , and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">$DKB$ bisects angle $ODC$ .</span></div></li></ol>

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

<p>In the diagram, $O$ is the centre of the smaller circle. $ABC$ and $AED$ 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/1770116694011-dxm0ek5ftd.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTY2OTQwMTEtZHhtMGVrNWZ0ZC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjkzNDYxLCJleHAiOjE3ODk2OTM3NjF9.-XUcYv5DqhuPmOwnIXFY7cq8LCpZBLLfS2PRMc0lI-M" style="max-width: min(230px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">$AC = AD$, and</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">$BE$ is parallel to $CD$.</span></div></li></ol>

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

<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="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770116565265-ro30kpxh2a.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTY1NjUyNjUtcm8zMGtweGgyYS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjkzNDYxLCJleHAiOjE3ODk2OTM3NjF9.F1Ve_iuACd8L8smf6QXH4Ng_SWBnXCl0tJJmFsNfMrg" style="max-width: min(270px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">triangles $ABT$ and $CAT$ are similar,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangles $AMS$ and $CAB$ are similar, and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">quadrilateral $ASOC$ is a trapezium.</span></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

<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>
<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/1770116438503-b85oj6t7sbp.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTY0Mzg1MDMtYjg1b2o2dDdzYnAucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.tNEv_YhHIuvauKqL2npGbg4ESgPhkfXMDfKLrhxhFRM" style="max-width: min(230px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $ODC = 2 \times \text{angle}\hspace{0.5em} OBC$,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangle $COD$ is an isosceles triangle, and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">angle $DCO = \text{angle}\hspace{0.5em} CAT + 90^\circ$.</span></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="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770116267635-0pgcysozb5pm.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTYyNjc2MzUtMHBnY3lzb3piNXBtLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3ODk2OTM0NjEsImV4cCI6MTc4OTY5Mzc2MX0.1TxJtd_7f0IsckhHTvNpsU7Swyp33y4T8o-eV4dqafE" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Geometry diagram of a circle with tangents and parallel lines." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $BCT =$ angle $ACD$,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangles $BTC$ and $DAC$ are similar, and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">$AD = BC$.</span></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

<p>In the diagram, $O$ is the centre of the smaller circle that intersects the larger circle at $B$, $C$ and $O$. $ABD$ 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>
<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/1770116087035-jt0h12ytsl.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTYwODcwMzUtanQwaDEyeXRzbC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzg5NjkzNDYxLCJleHAiOjE3ODk2OTM3NjF9.Zxh99Wmj4pnvb_CDhCzLlOrGpgl-1an_FmsCHqkSZ9k" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $AEB = 180^\circ - 2 \times$ angle $BAC$ ,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangle $ABE$ is an isosceles triangle, and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">angle $DFE = 360^\circ - 3 \times$ angle $BAC$ .</span></div></li></ol>

In the diagram, $O$ is the centre of the smaller circle that intersects the larger circle at $B$, $C$ and $O$. $ABD$ 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, 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="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770115953224-wq13l6l315a.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTU5NTMyMjQtd3ExM2w2bDMxNWEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.9OuywaDltGzPLSy0coJ_uF7ZcjZkfPsLrg-XE02-12I" style="max-width: min(230px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></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, the smaller circle and the larger circle intersect at $B$ and $D$. Given that&nbsp;$ABC$ and $EDC$ are straight lines and that&nbsp;$BE$ is a diameter of the smaller circle, 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/1770115804094-zhkzgr536cn.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTU4MDQwOTQtemhremdyNTM2Y24ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.zG_eLNE3WTjAhuIkRBY27Dswqn7djJhHU3rZ0AUjB1I" style="max-width: min(230px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">$BC$ is a diameter of the larger circle, and</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangles $BDC$ and $EAC$ are similar.</span></div></li></ol>

In the diagram, the smaller circle and the larger circle intersect at $B$ and $D$. Given that $ABC$ and $EDC$ are straight lines and that $BE$ is a diameter of the smaller circle, prove that

$BC$ is a diameter of the larger circle, and

triangles $BDC$ and $EAC$ are similar.

<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="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770115686223-c7xwgkesjfg.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTU2ODYyMjMtYzd4d2drZXNqZmcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.-YcSFbEXNCQoKCXp935OGaCzZhSr9fy_4R8eq3qlvOw" style="max-width: min(170px, 100%); height: auto; display: block;" alt="Geometry diagram showing two intersecting circles with tangents and a transversal line." class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">$RP$ is parallel to $AQ$, and</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">triangles $RPM$ and $QAM$ are similar.</span></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>
<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/1770115502422-wwnaapxr05d.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzAxMTU1MDI0MjItd3duYWFweHIwNWQucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc4OTY5MzQ2MSwiZXhwIjoxNzg5NjkzNzYxfQ.eo9R2hu_7eqpUTjsg6pJbWqIYtQS6pGStFCel5aHxfM" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-content"><span class="list-item-text">angle $ADS =$ angle $TBC$ ,</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">angle $CBD = 90^\circ - 2 \times$ angle $SDA$ , and</span></div></li><li data-list-item-id="item-iii"><div class="list-item-content"><span class="list-item-text">triangles $SAD$ and $BCD$ are similar.</span></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.

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