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PSLE

PSLE Foundation Mathematics

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE0MjI4Nzc3NDIta3JlazM3NzRiemkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzOCwiZXhwIjoxNzkyNzI1OTM4fQ.OKgw8WRP_Q4NbcC-UnlCFea8IPo9GKjRLDLtJXnwVkw" 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

Alternate Segment Theorem

Angles in the Same Segment

Similarity

Angles in a Semicircle and the Converse

2024 CGSS P1 Q13

Proofs in Plane Geometry

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM4MjU0MTktN3FjNGdtaGg0YzUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzOCwiZXhwIjoxNzkyNzI1OTM4fQ.AU0FwRYhszEO8B-IStf3_FJ2BvWKm8iY8T0WObhhBXw" 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM2ODU1MzYtb3c0OHM2dmI4OC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NjM4LCJleHAiOjE3OTI3MjU5Mzh9.GVXSqrWX-oD2n7M5XpWYeS0wPG9DuMEd2lUXKVtJEMk" 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>Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$.</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;"><p>$CD$ is the altitude (height) on side $AB$.</p><p>Show that</p><ol class="i-list"><li><div class="list-item-paragraph">$CD^2 = AD \times BD$</div></li><li><div class="list-item-paragraph">$AC^2 = AD \times AB$, and</div></li><li><div class="list-item-paragraph">$BC^2 = BD \times AB$</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/1770798940228-a7m71oynkw.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTg5NDAyMjgtYTdtNzFveW5rdy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NjM4LCJleHAiOjE3OTI3MjU5Mzh9.DCfCvXBFy2MzEvliPdA-59_9CcSmIgVzkpPiPdsOf-g" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$.

IP2U8 HOT Q3 (Right Triangle Altitude Theorem)

Congruence and Similarity

<p>Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$. $CD$ is the altitude (height) on side $AB$.</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;"><p>Show that&nbsp;$AB^2 = BC^2 + AC^2$.</p></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/1770797619579-v2cr56ql3jq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTc2MTk1NzktdjJjcjU2cWwzanEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzOCwiZXhwIjoxNzkyNzI1OTM4fQ.bomfhkAUdTbKajp0Aq67NL9-wDvqpz1DmWrYQEavPuE" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Triangle $ABC$ is a right-angled triangle with angle $ACB = 90^\circ$. $CD$ is the altitude (height) on side $AB$.

IP2U8 Proof of the Pythagoras' Theorem

<p>Two cylindrical cans of milk powder of the same brand are geometrically similar to each other as shown.</p>
<p>The heights of both cans are $10$ cm and $14$ cm respectively.</p>
<p>Given that the radius of the base of can $A$ is $6$ cm, calculate the radius of the base of can $B$.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">Find the ratio of the volume of can $A$ to the volume of can $B$.</div></li></ol>
<p>The net weight of can $A$ is $205$ g.</p>
<ol class="i-list" start="2"><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Find the net weight of can $B$, correct to the nearest gram.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">If the cost of cans $A$ and $B$ are $\$8.40$ and $\$19.50$ respectively, which can of milk powder is of better value? Explain your answer clearly.</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/1770796810249-cj6kumya4y9.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTY4MTAyNDktY2o2a3VteWE0eTkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzOCwiZXhwIjoxNzkyNzI1OTM4fQ.u4Ro-8f9i2JJPf72wTyQYLN6rex5Beu27m62APTIjK4" style="max-width: min(350px, 100%); height: auto; display: block;" alt="Diagram showing two similar cylindrical cans, Can A with height 10 cm and Can B with height 14 cm." class="alignright custom-width" /></figure><div class="image-caption" style="text-align: right;">Diagram showing two similar cylindrical cans, Can A with height 10 cm and Can B with height 14 cm.</div></div>

Two cylindrical cans of milk powder of the same brand are geometrically similar to each other as shown.

The heights of both cans are $10$ cm and $14$ cm respectively.

Given that the radius of the base of can $A$ is $6$ cm, calculate the radius of the base of can $B$.

Find the ratio of the volume of can $A$ to the volume of can $B$.

The net weight of can $A$ is $205$ g.

Find the net weight of can $B$, correct to the nearest gram.

If the cost of cans $A$ and $B$ are $\$8.40$ and $\$19.50$ respectively, which can of milk powder is of better value? Explain your answer clearly.

Volume Ratios of Similar Solids

Comparing Unit Prices

IP2U8 Exercise Q11

Ratio, Rate and Speed

<p>The two containers $A$ and $B$ shown in the diagram are geometrically similar.</p>
<p>The height of one container is $6$ m and the other is $12$ m.</p>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">The top of the larger container has a circumference of $22$ m, find the circumference of the top of the smaller container.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">If the surface area of the smaller container is $12.5$ m$^2$, find the surface area of the bigger container.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">The cost of the fruit juice in the larger container $B$ is $\$32$, find the cost of the fruit juice in the smaller container if both containers are completely filled with fruit juice.</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/1770795829903-0kwn0qi6d1w.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTU4Mjk5MDMtMGt3bjBxaTZkMXcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzOCwiZXhwIjoxNzkyNzI1OTM4fQ.vdLm7lUfyalR5mrvZeFrgwYD5UwV9v0S8MsLG43O6R8" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Containers A and B" class="alignright custom-width" /></figure><div class="image-caption" style="text-align: right;">Containers A and B</div></div>
<p>For part (iii), assume that both containers are charged the same price per unit volume of fruit juice.</p>

The two containers $A$ and $B$ shown in the diagram are geometrically similar.

The height of one container is $6$ m and the other is $12$ m.

The top of the larger container has a circumference of $22$ m, find the circumference of the top of the smaller container.

If the surface area of the smaller container is $12.5$ m$^2$, find the surface area of the bigger container.

For part (iii), assume that both containers are charged the same price per unit volume of fruit juice.

Area Ratios of Similar Figures

IP2U8 Exercise Q10

<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;"><p>In the diagram, $PQ$ is parallel to $ST$ and $XY$ is parallel to $RT$.</p><p>Given that&nbsp;$PRT$, $QRXS$ and $SYT$ are straight lines.</p><ol class="i-list"><li><div class="list-item-paragraph">Prove that&nbsp;$\Delta PQR$ and $\Delta YSX$ are similar.</div></li><li><div class="list-item-paragraph">Given that&nbsp;$QR = 4$ cm, $RX = 2$ cm, $XS = 8$ cm and $XY = 16$ cm. Find the numerical value of</div><ol class="abc-list"><li><div class="list-item-paragraph">$\frac{QR}{SR}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{Area of}\hspace{0.5em} \Delta SXY}{\text{Area of}\hspace{0.5em} \Delta SXT}$,</div></li><li><div class="list-item-paragraph">$\frac{\text{Area of}\hspace{0.5em} \Delta PQR}{\text{Area of}\hspace{0.5em} \Delta TSR}$.</div></li></ol></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/1770793501078-dmbi0upuvuq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTM1MDEwNzgtZG1iaTB1cHV2dXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzOCwiZXhwIjoxNzkyNzI1OTM4fQ.tRkiEvauZK4vff7eyMie9NBskDiaS5OqLuwb5ufki2A" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>
<ol class="i-list" start="3"><li><div class="list-item-paragraph">Hence, calculate the area of $\Delta TSR$, given that the area of $\Delta PQR$ is $25$ cm$^2$.</div></li></ol>

Area Ratios of Triangles with Equal Heights

IP2U8 Exercise Q6

<p>In the triangle $PQR$, $PQ = 9$ cm, $PR = 15$ cm and $K$ is a point on the side $PR$ such that&nbsp;$\angle PQK = \angle PRQ$.</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">Show that&nbsp;$\Delta PQK$ is similar to $\Delta PRQ$.</div></li><li><div class="list-item-paragraph">Hence, find the length of $PK$.</div></li><li><div class="list-item-paragraph">Given that the area of $\Delta PQK$ is $19.4$ cm$^2$, calculate</div><ol class="abc-list"><li><div class="list-item-paragraph">the area of $\Delta PRQ$,</div></li><li><div class="list-item-paragraph">the area of $\Delta KRQ$.</div></li></ol></li></ol></div><div class="question-column" style="flex: 1; align-self: center;"><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/1770792543937-or3jqx9mo7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTI1NDM5Mzctb3IzanF4OW1vNy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NjM4LCJleHAiOjE3OTI3MjU5Mzh9.UMCE8JJcdLa18N-FyCgqZsZqS0ZAzmNRk6XKwfNb6S8" style="max-width: min(260px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the triangle $PQR$, $PQ = 9$ cm, $PR = 15$ cm and $K$ is a point on the side $PR$ such that $\angle PQK = \angle PRQ$.

IP2U8 Exercise Q5

<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 $TB$ 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 $AC$.

Properties of Special Triangles

<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>
<div class="graph-block" data-graph-config="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" data-width="217" data-height="151" data-proportional style="width:217px;height:151px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:217px;--graph-aspect:217 / 151;"></div>
<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>
<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

Properties of Special Quadrilaterals

<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>
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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 $BCT =$ 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, 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>
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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">$BC$ is a diameter of the larger circle, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles $BDC$ and $EAC$ are similar.</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>
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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">$RP$ 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, $TA$ 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

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