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<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjcxMjQ1NjYtMjM0MGJ4Zmd6MzYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.O5bADIY-tKucYCdDXMGE7HD9GKK4VN9pPWu6BcLjeho" 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

Angles in a Semicircle and the Converse

Proving Concyclicity

IP2U14 HOT Q1

Properties of Circle

Polygons

<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 \(E\).

Alternate Segment Theorem

Angles in the Same Segment

Similarity

Proofs in Plane Geometry

<p>In the diagram, \(O\) is the centre of the smaller circle. \(ABC\) and \(AED\) are straight lines. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="252" data-height="198" data-proportional style="width:252px;height:198px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:252px;--graph-aspect:252 / 198;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">\(AC = AD\), and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(BE\) is parallel to \(CD\).</div></li></ol>
<p>As shown in the diagram, \(A,B,O,D\) lie on one circle, and \(B,C,D,E\) lie on the circle with centre \(O\).</p>

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

As shown in the diagram, \(A,B,O,D\) lie on one circle, and \(B,C,D,E\) lie on the circle with centre \(O\).

Angles at the Centre and Circumference

Angles Formed by Two Parallel Lines and a Transversal

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

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

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

triangle \(ABE\) is an isosceles triangle, and

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

<p>In the diagram, 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>
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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, \(ABCD\) 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.

Tangents Perpendicular to Radii

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

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

triangle \(BAT\) is an isosceles triangle.

<p>In the diagram, \(TA\) is the tangent to the circle at \(A\) and is parallel to \(CB\). Given that&nbsp;\(TDC\) 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 \(ATD =\) angle \(BAD\), and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(BA \times TD = AT \times AD\).</div></li></ol>

In the diagram, \(TA\) is the tangent to the circle at \(A\) and is parallel to \(CB\). Given that \(TDC\) is a straight line, prove that

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

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

triangles \(CBD\) and \(CAE\) are similar, and

As shown in the diagram, \(A,B,D,E\) lie on a circle.

<p>In the diagram, \(ABCD\) is a cyclic quadrilateral with \(AD = BC\). Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="178" data-height="166" data-proportional style="width:178px;height:166px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:178px;--graph-aspect:178 / 166;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">\(BD = AC\), and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">quadrilateral \(ABCD\) is a trapezium.</div></li></ol>

In the diagram, \(ABCD\) is a cyclic quadrilateral with \(AD = BC\). Prove that

Properties of Special Quadrilaterals

<p>In the diagram, \(BCE\) and \(ADE\) are straight lines. Given that angle \(BAD =\) angle \(ACD\), prove that</p>
<div class="graph-block" data-graph-config="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" data-width="254" data-height="142" data-proportional style="width:254px;height:142px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:254px;--graph-aspect:254 / 142;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">\(BD = AD\), and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(CD\) bisects angle \(ACE\).</div></li></ol>

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

<p>In the figure, two circles intersect at \(B\) and \(E\). \(ABC\) and \(DEF\) are straight lines. Prove that&nbsp;\(AF\) is parallel to \(CD\).</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/1768111433813-wtjqcfh6i7i.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE0MzM4MTMtd3RqcWNmaDZpN2kucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.ollHwsY6gI2Rm3BE7iyVYqxSIVPsSFwIdXoSoRAyRTc" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the figure, two circles intersect at \(B\) and \(E\). \(ABC\) and \(DEF\) are straight lines. Prove that \(AF\) is parallel to \(CD\).

<p>In the figure, \(O\) is the centre of the circle \(BCDE\) and \(ABC\) and \(AED\) are straight lines, prove that</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list"><li data-list-item-id="item-1780398684393-mvhu24dmu" data-list-item-path="item-1780398684393-mvhu24dmu"><div class="list-item-paragraph">\(AD=AC\),</div></li><li data-list-item-id="item-1780398684841-fkn4cdoq8" data-list-item-path="item-1780398684841-fkn4cdoq8"><div class="list-item-paragraph">Given that&nbsp;\(AB=AE\), show that&nbsp;\(BE\) is parallel to \(CD\).</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768111313403-sdv4ekep96g.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTEzMTM0MDMtc2R2NGVrZXA5NmcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.lciterQZh9VTx1euYuYadaJqSxjtxuWIaXVNpr92gqY" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>
<p>As shown in the diagram, \(A,B,O,D\) lie on one circle, and \(B,C,D,E\) lie on the circle with centre \(O\).</p>

In the figure, \(O\) is the centre of the circle \(BCDE\) and \(ABC\) and \(AED\) are straight lines, prove that

<p>In the figure, \(TA\) is a tangent to the circle at \(A\) and is parallel to \(CB\). Given that&nbsp;\(TDC\) is a straight line, prove that</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list"><li><div class="list-item-paragraph">\(\angle ATD = \angle BAD\),</div></li><li><div class="list-item-paragraph">\(BA \times TD = AT \times AD\).</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/1768110189625-y1cpw9vtna.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTAxODk2MjUteTFjcHc5dnRuYS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzQ0LCJleHAiOjE3OTI3MjYwNDR9.EgmC3KkZAZt94TyQqlSCrPsP6mgWLt8Rs3d5-vtW_uc" style="max-width: min(180px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(TA\) is a tangent to the circle at \(A\) and is parallel to \(CB\). Given that \(TDC\) is a straight line, prove that

<p>In the figure, \(O\) is the centre of the circle \(ABCD\) and it intersects the circle \(AOCE\) at \(A\) and \(C\). Given that&nbsp;\(ECB\) and \(ADE\) are straight lines, prove that</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list"><li><div class="list-item-paragraph">\(DE = CE\),</div></li><li><div class="list-item-paragraph">\(\angle CDE = \angle BAD\).</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/1768107234018-pwb2bh37fue.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDcyMzQwMTgtcHdiMmJoMzdmdWUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.AfsfQ2GcckWGvNBzt2qM0lDN1IMlSorAp4KSgmo87Gw" style="max-width: min(210px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(O\) is the centre of the circle \(ABCD\) and it intersects the circle \(AOCE\) at \(A\) and \(C\). Given that \(ECB\) and \(ADE\) are straight lines, prove that

<p>In the figure, \(BCE\) and \(ADE\) are straight lines. Given that&nbsp;\(\angle BAD = \angle ACD\), prove that</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="i-list"><li><div class="list-item-paragraph">\(CD\) bisects \(\angle ACE\),</div></li><li><div class="list-item-paragraph">\(BD = AD\).</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/1775564392342-rnyt2skns0g.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzU1NjQzOTIzNDItcm55dDJza25zMGcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.nbStrg2_J6heCCKCTNmgT3LJCDmzYiGPEJVOJeeKJog" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(BCE\) and \(ADE\) are straight lines. Given that \(\angle BAD = \angle ACD\), prove that

<p>In the given diagram, triangle \(\Delta ABC\) is an isosceles triangle with \(BA=BC\). Prove that&nbsp;\(AD=CE\).</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/1768106814451-ft1693r0nuw.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDY4MTQ0NTEtZnQxNjkzcjBudXcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.bKc-CV2e9e4mVhLqP9SJvffmsPtQsQ9Umyr_pmlz3I8" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the given diagram, triangle \(\Delta ABC\) is an isosceles triangle with \(BA=BC\). Prove that \(AD=CE\).

Recap of Angle Properties of Circles

<p>In the diagram, \(A\), \(B\), \(C\) and \(D\) are points on the circumference of the circle centre \(O\).<br>\(ABE\) and \(DCE\) are straight lines and \(ST\) is a tangent to the circle at \(A\).<br>\(ST\) is parallel to \(DE\), \(\angle CEB=36{}^\circ\) and \(\angle DAB=83{}^\circ\).</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;"><ol class="abc-list"><li><div class="list-item-paragraph">Calculate</div><ol class="i-list"><li><div class="list-item-paragraph">angle \(OAB\),</div></li><li><div class="list-item-paragraph">angle \(ABC\),</div></li><li><div class="list-item-paragraph">angle \(AOD\).</div></li></ol></li><li><div class="list-item-paragraph">Show that&nbsp;\(ACD\) is an isosceles triangle.</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/1767601914948-dxhsqkds96b.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDE5MTQ5NDgtZHhoc3FrZHM5NmIucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.ZdP_fNAMoV5I9w7_r9x_38VPaInqAhHOXUIWFjG76OY" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, \(A\), \(B\), \(C\) and \(D\) are points on the circumference of the circle centre \(O\).
\(ABE\) and \(DCE\) are straight lines and \(ST\) is a tangent to the circle at \(A\).
\(ST\) is parallel to \(DE\), \(\angle CEB=36{}^\circ\) and \(\angle DAB=83{}^\circ\).

Investigating Geometric Proofs through Circle Angle Invariants

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDE2MjcyMDktdTF3bW9yanV1d28ucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0NCwiZXhwIjoxNzkyNzI2MDQ0fQ.9F0OM42zc6hxJ5skLYuzmaxeqSZk_LmmzhC80slk-UI" 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,

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