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PSLE

PSLE Foundation Mathematics

<p>Triangle $ABC$ has&nbsp;$AB=26$ cm, $BC=10$ cm, and $AC=24$ cm. A circle passes through $A,B,C$.</p>
<div class="graph-block" data-graph-config="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" data-width="350" data-height="250" style="width:350px;height:250px;max-width:100%;display:block;margin:0 auto;"></div>
<p>Explain, with reasons, why $AB$ is the diameter of the circle.<span class="inline-marks">[3]</span></p>

Triangle $ABC$ has $AB=26$ cm, $BC=10$ cm, and $AC=24$ cm. A circle passes through $A,B,C$.

Explain, with reasons, why $AB$ is the diameter of the circle.

The Converse of Pythagoras Theorem

Angles in a Semicircle and the Converse

2025 YCKSS PRELIMS P1 Q12

Properties of Circle

Pythagoras’ Theorem

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE0MjI4Nzc3NDIta3JlazM3NzRiemkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.yNJLkjOnVvixrOEe6lrN6qNowpeY0fF0ihloH79S9tk" 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

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM4MjU0MTktN3FjNGdtaGg0YzUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.Lonfii89pVNDtP1Eyx_bhD55_5hoBVJyu0LYRDnkQSA" 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>$AB$, $AC$ and $AD$ are chords of a circle such that&nbsp;$AB = 4$ cm, $AC = 10$ cm and $AD = 11$ cm. Given that&nbsp;$AC$ bisects $\angle BAD$, find the area of the circle.</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/1770728619550-uu6wi9exw8l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3Mjg2MTk1NTAtdXU2d2k5ZXh3OGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.pCg_dQTshAZujfo4CV82QZpkjvkwATxKECE7RT5JBEU" style="max-width: min(160px, 100%); height: auto; display: block;" alt="Chords in a circle" class="alignright custom-width" /></figure></div>

$AB$, $AC$ and $AD$ are chords of a circle such that $AB = 4$ cm, $AC = 10$ cm and $AD = 11$ cm. Given that $AC$ bisects $\angle BAD$, find the area of the circle.

Equal Chords and Subtended Angles

Cosine Rule

Circumference and Area of Circles

IP2U14 HOT Q5

Trigonometry

Perimeter and Area of Plane Figures

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjcxMjQ1NjYtMjM0MGJ4Zmd6MzYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.CHK0Bc9WXSotasEGDyRLz_YGm1gBtAhzbSuEpt-o86E" 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

Properties of Special Triangles

Proving Concyclicity

IP2U14 HOT Q1

Polygons

<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="graph-block" data-graph-config="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" data-width="227" data-height="231" data-proportional style="width:227px;height:231px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:227px;--graph-aspect:227 / 231;"></div>
<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, $TA$ 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, 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="graph-block" data-graph-config="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" data-width="255" data-height="154" data-proportional style="width:255px;height:154px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:255px;--graph-aspect:255 / 154;"></div>
<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, $SA$ 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, $PT$ is the tangent to the semicircle with centre $O$ at $P$ and $BP$ bisects angle $ABT$. 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 $BTP$ and $BPA$ are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">angle $BTP = 90^{\circ}$.</div></li></ol>

In the diagram, $PT$ is the tangent to the semicircle with centre $O$ at $P$ and $BP$ bisects angle $ABT$. Prove that

triangles $BTP$ and $BPA$ are similar, and

<p>In the diagram, $O$ is the centre of the circle, $AC$ is a diameter and $PAQ$ is the tangent to the circle at $A$. Prove that</p>
<div class="graph-block" data-graph-config="eyJib3VuZGluZ0JveCI6Wy0wLjEwMjMsNy43NCw5Ljg1MzcsMC4wMjcyXSwid2lkdGgiOjIyMywiaGVpZ2h0IjoxNzMsImF4aXMiOmZhbHNlLCJncmlkIjpmYWxzZSwiZ3JpZFNldHRpbmdzIjp7Im1ham9yU3RlcFgiOjEsIm1ham9yU3RlcFkiOjEsIm1pbm9yRWxlbWVudHNYIjo0LCJtaW5vckVsZW1lbnRzWSI6NCwic2hvd01ham9yR3JpZExpbmVzIjp0cnVlLCJzaG93TWlub3JHcmlkTGluZXMiOnRydWUsIm1ham9yQ29sb3IiOiIjOTRhM2I4IiwibWFqb3JPcGFjaXR5IjowLjQ1LCJtYWpvclN0cm9rZVdpZHRoIjowLjgsIm1pbm9yQ29sb3IiOiIjY2JkNWUxIiwibWlub3JPcGFjaXR5IjowLjQ1LCJtaW5vclN0cm9rZVdpZHRoIjowLjV9LCJrZWVwQXNwZWN0UmF0aW8iOnRydWUsInNob3dOYXZpZ2F0aW9uIjpmYWxzZSwic2hvd0F4aXNMYWJlbHMiOmZhbHNlLCJ4VGlja0xhYmVsRXZlcnkiOjEsInlUaWNrTGFiZWxFdmVyeSI6MSwieEF4aXNMYWJlbCI6IlxcKHhcXCkiLCJ5QXhpc0xhYmVsIjoiXFwoeVxcKSIsIm9yaWdpbkxhYmVsIjoiXFwoT1xcKSIsInhBeGlzTGFiZWxPZmZzZXQiOlsyLC0xMV0sInlBeGlzTGFiZWxPZmZzZXQiOlstMTQsNV0sIm9yaWdpbkxhYmVsT2Zmc2V0IjpbLTUsLTJdLCJlbGVtZW50cyI6W3siaWQiOiJpbXBvcnRlZC1jaXJjbGUtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtMCIsInR5cGUiOiJjaXJjbGUiLCJjZW50ZXIiOlszLjU4LDMuNzldLCJyYWRpdXMiOjMuMDIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjUsImZpbGxDb2xvciI6Im5vbmUifSx7ImlkIjoiaW1wb3J0ZWQtc2VnbWVudC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC0xIiwidHlwZSI6InNlZ21lbnQiLCJwb2ludDEiOlswLjgxLDAuNzddLCJwb2ludDIiOlszLjU4LDYuODFdLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtc2VnbWVudC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC0yIiwidHlwZSI6InNlZ21lbnQiLCJwb2ludDEiOlszLjU4LDYuODFdLCJwb2ludDIiOls5LjIsMC43N10sImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC1zZWdtZW50LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTMiLCJ0eXBlIjoic2VnbWVudCIsInBvaW50MSI6WzAuODEsMC43N10sInBvaW50MiI6WzkuMiwwLjc3XSwiY29sb3IiOiIjMDAwMDAwIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXNlZ21lbnQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtNCIsInR5cGUiOiJzZWdtZW50IiwicG9pbnQxIjpbMy41OCw2LjgxXSwicG9pbnQyIjpbMy41OCwwLjc3XSwiY29sb3IiOiIjMDAwMDAwIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXNlZ21lbnQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtNSIsInR5cGUiOiJzZWdtZW50IiwicG9pbnQxIjpbMS4yOTEzNTQsMS44MTk1OTRdLCJwb2ludDIiOlszLjU4LDAuNzddLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtc2VnbWVudC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC02IiwidHlwZSI6InNlZ21lbnQiLCJwb2ludDEiOlszLjU4LDAuNzddLCJwb2ludDIiOls2LjU5MjE3MywzLjU3MjcxOF0sImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC1wb2ludC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC03IiwidHlwZSI6InBvaW50IiwieCI6My41OCwieSI6My43OSwicG9pbnRTaXplIjoxLCJwb2ludFN0eWxlIjoiY2lyY2xlIiwibGFiZWwiOiIiLCJjb2xvciI6IiMwMDAwMDAiLCJmaWxsQ29sb3IiOiIjMDAwMDAwIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXRleHQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtOCIsInR5cGUiOiJ0ZXh0IiwieCI6My4zNCwieSI6MC4zLCJsYWJlbCI6IlxcKEFcXCkiLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtdGV4dC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC05IiwidHlwZSI6InRleHQiLCJ4IjowLjU3LCJ5IjoxLjYyLCJsYWJlbCI6IlxcKEJcXCkiLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtdGV4dC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC0xMCIsInR5cGUiOiJ0ZXh0IiwieCI6My4zMywieSI6Ny4yNCwibGFiZWwiOiJcXChDXFwpIiwiY29sb3IiOiIjMDAwMDAwIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXRleHQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtMTEiLCJ0eXBlIjoidGV4dCIsIngiOjYuNzgsInkiOjMuODksImxhYmVsIjoiXFwoRFxcKSIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC10ZXh0LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTEyIiwidHlwZSI6InRleHQiLCJ4IjozLjksInkiOjMuNzIsImxhYmVsIjoiXFwoT1xcKSIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC10ZXh0LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTEzIiwidHlwZSI6InRleHQiLCJ4IjowLjE1LCJ5IjowLjMsImxhYmVsIjoiXFwoUFxcKSIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC10ZXh0LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTE0IiwidHlwZSI6InRleHQiLCJ4Ijo5LjMsInkiOjAuMzYsImxhYmVsIjoiXFwoUVxcKSIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9XSwiYWxpZ25tZW50IjoiY2VudGVyIn0=" data-width="223" data-height="173" data-proportional style="width:223px;height:173px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:223px;--graph-aspect:223 / 173;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles $CAP$ and $CBA$ are similar, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AC^2 = CP \times CB$.</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">State a triangle that is similar to triangle $CAQ$. Hence, or otherwise, prove that&nbsp;$CP \times CB = CD \times CQ$.</div></li></ol>

In the diagram, $O$ is the centre of the circle, $AC$ is a diameter and $PAQ$ is the tangent to the circle at $A$. Prove that

triangles $CAP$ and $CBA$ are similar, and

State a triangle that is similar to triangle $CAQ$. Hence, or otherwise, prove that $CP \times CB = CD \times CQ$.

<p>In the figure, $O$ is the centre of the circle, $AC$ is a diameter and $PAQ$ is a tangent to the circle at $A$.</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;$\Delta CAP$ is similar to $\Delta CBA$.</div></li><li><div class="list-item-paragraph">Prove that&nbsp;$AC^2=CP \times CB$.</div></li><li><div class="list-item-paragraph">State a triangle that is similar to $\Delta CAQ$.</div></li></ol><p>Hence, or otherwise, prove that&nbsp;$CP \times CB = CD \times CQ$.</p></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/1768112050469-d7q2dds3l0h.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTIwNTA0NjktZDdxMmRkczNsMGgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.mpTO1ISazuS7eshpQzE0OwABya5FFCou50lP693AAK4" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, $O$ is the centre of the circle, $AC$ is a diameter and $PAQ$ is a tangent to the circle at $A$.

<p>In the figure, the smaller circle $ADE$ intersects the larger circle $ABCD$ at $A$ and $D$. Given that&nbsp;$EAB$ and $EDC$ are straight lines and that&nbsp;$AC$ is the diameter of the larger circle $ABCD$, prove that&nbsp;$AE$ is the diameter of the smaller circle.</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/1768111580793-yidrjhuno1.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE1ODA3OTMteWlkcmpodW5vMS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NTc1LCJleHAiOjE3OTI3MjU4NzV9.c6p3gX7CMLnTp7CQkMXH_UuaDicEMi2WJU7BFP-BlWc" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the figure, the smaller circle $ADE$ intersects the larger circle $ABCD$ at $A$ and $D$. Given that $EAB$ and $EDC$ are straight lines and that $AC$ is the diameter of the larger circle $ABCD$, prove that $AE$ is the diameter of the smaller circle.

<p>In the figure, the smaller circle $ABDE$ intersects the larger circle $BCD$ at $B$ and $D$. Given that&nbsp;$ABC$ and $EDC$ are straight lines and that&nbsp;$BE$ is the diameter of the smaller circle $ABDE$, prove that&nbsp;$BC$ is the diameter of the larger circle.</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/1768111504182-7fhpkrfxc5f.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE1MDQxODItN2ZocGtyZnhjNWYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.n6v2fE_84kGrNQSOzpnhNL1lY7Y-N0nLE7gi75k06DY" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the figure, the smaller circle $ABDE$ intersects the larger circle $BCD$ at $B$ and $D$. Given that $ABC$ and $EDC$ are straight lines and that $BE$ is the diameter of the smaller circle $ABDE$, prove that $BC$ is the diameter of the larger circle.

<p>In the diagram, $PT$ is a tangent to the semicircle $APB$. If&nbsp;$PB$ bisects $\angle ABT$, prove that&nbsp;$\angle BTP=90{}^\circ $.</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/1768110913020-lbddcwd9orn.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA5MTMwMjAtbGJkZGN3ZDlvcm4ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.o9QXtLw0VqyNLEjP1WTQTP6X_2-eUYcGTba3A1O-RTs" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram, $PT$ is a tangent to the semicircle $APB$. If $PB$ bisects $\angle ABT$, prove that $\angle BTP=90{}^\circ $.

Triangle Angle Sum and Exterior Angles

<p>In the diagram, $AC$ is a diameter of the circle with centre $O$. $ACD$ is a straight line and $AB=BD$. Prove that</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li data-list-item-id="item-1780398315636-ff57u9wfv" data-list-item-path="item-1780398315636-ff57u9wfv"><div class="list-item-paragraph">$BC=DC$,</div></li><li data-list-item-id="item-1780398316692-o1non6tqr" data-list-item-path="item-1780398316692-o1non6tqr"><div class="list-item-paragraph">$\angle BAC=30{}^\circ $.</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768110395556-2169c09d1se.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTAzOTU1NTYtMjE2OWMwOWQxc2UucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.6QoD4Hq-kkRMqD-ssJ7xZvk8FLidZCadmr8_ZuO-cWg" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>
<p>$BD$ is tangent to the circle at $B$.</p>

In the diagram, $AC$ is a diameter of the circle with centre $O$. $ACD$ is a straight line and $AB=BD$. Prove that

$BD$ is tangent to the circle at $B$.

<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 below, $O$ is the centre of the circle and points $A$, $B$, $C$ and $D$ lie on the circumference of the circle.<br>The tangent at $A$ meets $CD$ produced at $E$.<br>The line $AC$ passes through $O$ and lines $AC$ intersects $BD$ at point $F$.<br>It is also given that&nbsp;$\angle ACD=46{}^\circ$ and $\angle CAB=25{}^\circ$.<br>Stating your reasons clearly, find</p><ol class="abc-list"><li><div class="list-item-paragraph">$\angle ADO$.</div></li><li><div class="list-item-paragraph">$\angle OFD$.</div></li><li><div class="list-item-paragraph">$\angle AEC$.</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/1767602520657-fzpds21ague.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDI1MjA2NTctZnpwZHMyMWFndWUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.PpUW7CDGOITGXE6vn-r4bas-daOOda4YrWCRITYcUlU" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

The Geometry of Inscribed Triangles and Cyclic Polygons

<p>In the diagram, $O$ is the centre of the circle through $A$, $B$, $C$ and $D$ and $TB$ is the tangent to the circle at $B$.<br>The diameter $AC$ and the chord $BD$ intersects at $X$.<br>Given&nbsp;$\angle DCX = 32{}^\circ$ and $\angle OBC = 40{}^\circ$, calculate, stating your reasons clearly,</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">$\angle OAB$,</div></li><li><div class="list-item-paragraph">$\angle CXD$,</div></li><li><div class="list-item-paragraph">$\angle ABT$,</div></li><li><div class="list-item-paragraph">$\angle BAD$.</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/1767601627209-u1wmorjuuwo.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDE2MjcyMDktdTF3bW9yanV1d28ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.wAwxmdnr4Op8JWAobZw__znuC2yDmqs405WDH1wFSms" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, $O$ is the centre of the circle through $A$, $B$, $C$ and $D$ and $TB$ is the tangent to the circle at $B$.
The diameter $AC$ and the chord $BD$ intersects at $X$.
Given $\angle DCX = 32{}^\circ$ and $\angle OBC = 40{}^\circ$, calculate, stating your reasons clearly,

Euclidean Geometry: Interior Angles and Circular Arc Relationships

<p>In the circle with centre $O$, $AD$ is a diameter.</p>
<p>The points $B$ and $C$ lie on the circumference of the circle such that&nbsp;$\angle ADB = 28^\circ$ and $\angle BDC = 42^\circ$.</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>Given that&nbsp;$BCE$ is a straight line, find</p><ol class="abc-list"><li><div class="list-item-paragraph">$\angle ABD$,</div></li><li><div class="list-item-paragraph">$\angle BAC$,</div></li><li><div class="list-item-paragraph">$\angle DCE$.</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/1767601123722-sf58xhxg18h.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDExMjM3MjItc2Y1OHhoeGcxOGgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU3NSwiZXhwIjoxNzkyNzI1ODc1fQ.dhgYxOEEGaPOXHeVu-SDYy5HQ6J4Ih9AqE_i7po9KaQ" style="max-width: min(220px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the circle with centre $O$, $AD$ is a diameter.

The points $B$ and $C$ lie on the circumference of the circle such that $\angle ADB = 28^\circ$ and $\angle BDC = 42^\circ$.

Angles in the Same Segment of a Circle: Equality Proofs

<p>Find the value of all the unknowns in each diagram, stating your reasons clearly.</p>
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Find the value of all the unknowns in each diagram, stating your reasons clearly.

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