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PSLE

PSLE Foundation Mathematics

<p>Points $A$, $B$, $C$ and $D$ lie on a circle. The tangent to the circle at $A$ meets $BD$ produced at $E$. $AE$ is parallel to $BC$.</p>
<div class="graph-block" data-graph-config="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" data-width="300" data-height="200" style="width:300px;height:200px;max-width:100%;display:block;margin:0 auto;"></div>
<p>Prove that</p>
<ol class="i-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">$AB = AC$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">$\angle CDE = 2\angle ABC$.<span class="inline-marks">[3]</span></div></li></ol>

Points $A$, $B$, $C$ and $D$ lie on a circle. The tangent to the circle at $A$ meets $BD$ produced at $E$. $AE$ is parallel to $BC$.

Alternate Segment Theorem

Properties of Special Triangles

Angles in the Same Segment

2024 FMSS PRELIMS P1 Q3

Proofs in Plane Geometry

<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1781422877742-krek3774bzi.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODE0MjI4Nzc3NDIta3JlazM3NzRiemkucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.yi7jYQh_Nq0d0G-g-oWeFwoIbJgq9ps1CMH955FCziI" style="max-width: min(400px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<p>The diagram shows a circle, centre $O$. Points $A$, $B$, $D$ and $E$ lie on the circle. <br>The line $AD$ is a diameter of the circle and $DCP$ is the tangent of the circle at $D$. <br>The lines $EB$ and $AD$ intersect at point $Q$. <br>The chord $AB$ is produced to meet the tangent $DCP$ at $P$. <br>It is given that&nbsp;$\text{angle}\hspace{0.5em} CBP = \text{angle}\hspace{0.5em} DEB$.</p>
<ol class="abc-list"><li data-list-item-id="item-1" data-list-item-path="item-1"><div class="list-item-paragraph">Show that&nbsp;$\text{angle}\hspace{0.5em} BCD = 90^\circ$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-2" data-list-item-path="item-2"><div class="list-item-paragraph">Show that&nbsp;$DP^2 = BP^2 + DA \times BC$.<span class="inline-marks">[4]</span></div></li></ol>

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

Applications of Pythagoras’ Theorem

Similarity

Angles in a Semicircle and the Converse

2024 CGSS P1 Q13

<p>In the diagram, $PQS$ 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM2ODU1MzYtb3c0OHM2dmI4OC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NTQwLCJleHAiOjE3OTI3MjU4NDB9.QRYTFe3yjo0PIEg2lKRnf9B475CDNsi5zmTHERewnIo" 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>$AB$, $AC$ 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3Mjg2MTk1NTAtdXU2d2k5ZXh3OGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.81pXVn8ttZboQv5-2XYpok3Hgu2o3OqLbvKOHz-E1qs" 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

Properties of Circle

Trigonometry

Perimeter and Area of Plane Figures

<p>$ABCD$ is a convex quadrilateral such that&nbsp;$\angle BAD=56^\circ$ and $\angle ADC=108^\circ$. $DC$ is produced to $P$ such that&nbsp;$\angle BPD=52^\circ$ and $BC$ bisects $\angle ABP$. If&nbsp;$\angle BDC=35^\circ$, find $\angle ACB$ and $\angle ACP$.</p>

$ABCD$ is a convex quadrilateral such that $\angle BAD=56^\circ$ and $\angle ADC=108^\circ$. $DC$ is produced to $P$ such that $\angle BPD=52^\circ$ and $BC$ bisects $\angle ABP$. If $\angle BDC=35^\circ$, find $\angle ACB$ and $\angle ACP$.

Proving Concyclicity

IP2U14 HOT Q3

<p>Chords $AC$ and $BD$, perpendicular to each other, intersect at $E$.</p>
<p>In $\Delta ADE$, the altitude from $E$ meets $AD$ at $F$. $FE$ produced meets $BC$ at $G$. Prove that&nbsp;$G$ is the midpoint of $BC$.</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/1770726144466-1u5dr35xwkc.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjYxNDQ0NjYtMXU1ZHIzNXh3a2MucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.A88M0tfgArkOxMZusd5mHm6j9XCIzWLLZlJuVW3kgNg" style="max-width: min(170px, 100%); height: auto; display: block;" alt="Geometric construction of perpendicular chords and an altitude within a circle" class="alignright custom-width" /></figure></div>

Chords $AC$ and $BD$, perpendicular to each other, intersect at $E$.

In $\Delta ADE$, the altitude from $E$ meets $AD$ at $F$. $FE$ produced meets $BC$ at $G$. Prove that $G$ is the midpoint of $BC$.

IP2U14 Exercise Q3

Polygons

<p>$A$, $B$, $C$, $D$ and $E$ are points on the circle such that&nbsp;$AC$ is perpendicular to $BE$ and $EC$ is perpendicular to $AD$. Prove that&nbsp;$\angle BCD$ is twice of $\angle ACE$.</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/1770725840886-skhnb16n6x8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjU4NDA4ODYtc2tobmIxNm42eDgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.PsBOoilHtsbzghx7ldG9bHuMLLJR2yT9JlrxBJXUi3M" style="max-width: min(160px, 100%); height: auto; display: block;" alt="Points A, B, C, D, and E on a circle with intersecting chords AC, BE and EC, AD." class="alignright custom-width" /></figure></div>

$A$, $B$, $C$, $D$ and $E$ are points on the circle such that $AC$ is perpendicular to $BE$ and $EC$ is perpendicular to $AD$. Prove that $\angle BCD$ is twice of $\angle ACE$.

IP2U14 Exercise Q2

<p>$A$, $M$, $B$, $C$ and $N$ are points on a circle. $BN$ and $MC$, the angle bisectors of $\angle ABC$ and $\angle ACB$ respectively, intersect at $P$. Prove that the point $N$ is equidistant from $A$ and $P$.</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/1770724970454-60j7w6kdkqe.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjQ5NzA0NTQtNjBqN3c2a2RrcWUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.2eLvnF5bxuP_5JPMiY-Ne_-g07m2JIjDhz2tK81es1U" style="max-width: min(150px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

$A$, $M$, $B$, $C$ and $N$ are points on a circle. $BN$ and $MC$, the angle bisectors of $\angle ABC$ and $\angle ACB$ respectively, intersect at $P$. Prove that the point $N$ is equidistant from $A$ and $P$.

Congruence

IP2U14 Exercise Q1

<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

<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 $AC$.

Equal Tangents from an External Point

<p>In the diagram, $O$ is the centre of the larger circle. $AOB$, $AKC$ and $BKD$ 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">$DK = CK$ ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">$AB$ is parallel to $CD$ , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">$DKB$ bisects angle $ODC$ .</div></li></ol>
<p>As shown, $A,B,C,D$ lie on the larger circle and $A,O,D,K$ lie on the smaller circle.</p>

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

As shown, $A,B,C,D$ lie on the larger circle and $A,O,D,K$ lie on the smaller circle.

Angles at the Centre and Circumference

<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

<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, $AP$ 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 with $AD = BC$. Prove that</p>
<div class="graph-block" data-graph-config="eyJib3VuZGluZ0JveCI6Wy0wLjA4MSw3LjI3NzcsNy42NTg4LDAuMDY3OF0sIndpZHRoIjoxNzgsImhlaWdodCI6MTY2LCJheGlzIjpmYWxzZSwiZ3JpZCI6ZmFsc2UsImdyaWRTZXR0aW5ncyI6eyJtYWpvclN0ZXBYIjoxLCJtYWpvclN0ZXBZIjoxLCJtaW5vckVsZW1lbnRzWCI6NCwibWlub3JFbGVtZW50c1kiOjQsInNob3dNYWpvckdyaWRMaW5lcyI6dHJ1ZSwic2hvd01pbm9yR3JpZExpbmVzIjp0cnVlLCJtYWpvckNvbG9yIjoiIzk0YTNiOCIsIm1ham9yT3BhY2l0eSI6MC40NSwibWFqb3JTdHJva2VXaWR0aCI6MC44LCJtaW5vckNvbG9yIjoiI2NiZDVlMSIsIm1pbm9yT3BhY2l0eSI6MC40NSwibWlub3JTdHJva2VXaWR0aCI6MC41fSwia2VlcEFzcGVjdFJhdGlvIjp0cnVlLCJzaG93TmF2aWdhdGlvbiI6ZmFsc2UsInNob3dBeGlzTGFiZWxzIjpmYWxzZSwieFRpY2tMYWJlbEV2ZXJ5IjoxLCJ5VGlja0xhYmVsRXZlcnkiOjEsInhBeGlzTGFiZWwiOiJcXCh4XFwpIiwieUF4aXNMYWJlbCI6IlxcKHlcXCkiLCJvcmlnaW5MYWJlbCI6IlxcKE9cXCkiLCJ4QXhpc0xhYmVsT2Zmc2V0IjpbMiwtMTFdLCJ5QXhpc0xhYmVsT2Zmc2V0IjpbLTE0LDVdLCJvcmlnaW5MYWJlbE9mZnNldCI6Wy01LC0yXSwiZWxlbWVudHMiOlt7ImlkIjoiaW1wb3J0ZWQtY2lyY2xlLWJsb2NrLTAtY29udGVudC1lbGVtZW50LTAiLCJ0eXBlIjoiY2lyY2xlIiwiY2VudGVyIjpbNC4wMiwzLjMzXSwicmFkaXVzIjozLjE5LCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41LCJmaWxsQ29sb3IiOiJub25lIn0seyJpZCI6ImltcG9ydGVkLXNlZ21lbnQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtMSIsInR5cGUiOiJzZWdtZW50IiwicG9pbnQxIjpbMC44MzYxNzQsMy4xMzE2MzFdLCJwb2ludDIiOls2Ljk5ODkzMSwyLjE4ODkxOF0sImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC1zZWdtZW50LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTIiLCJ0eXBlIjoic2VnbWVudCIsInBvaW50MSI6WzYuOTk4OTMxLDIuMTg4OTE4XSwicG9pbnQyIjpbNS42NzYwNjcsNi4wNTY0NTJdLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtc2VnbWVudC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC0zIiwidHlwZSI6InNlZ21lbnQiLCJwb2ludDEiOls1LjY3NjA2Nyw2LjA1NjQ1Ml0sInBvaW50MiI6WzMuMDA4MTk5LDYuMzU1Mjg2XSwiY29sb3IiOiIjMDAwMDAwIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXNlZ21lbnQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtNCIsInR5cGUiOiJzZWdtZW50IiwicG9pbnQxIjpbMy4wMDgxOTksNi4zNTUyODZdLCJwb2ludDIiOlswLjgzNjE3NCwzLjEzMTYzMV0sImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC1zZWdtZW50LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTUiLCJ0eXBlIjoic2VnbWVudCIsInBvaW50MSI6WzAuODM2MTc0LDMuMTMxNjMxXSwicG9pbnQyIjpbNS42NzYwNjcsNi4wNTY0NTJdLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtc2VnbWVudC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC02IiwidHlwZSI6InNlZ21lbnQiLCJwb2ludDEiOlszLjAwODE5OSw2LjM1NTI4Nl0sInBvaW50MiI6WzYuOTk4OTMxLDIuMTg4OTE4XSwiY29sb3IiOiIjMDAwMDAwIiwic3Ryb2tlV2lkdGgiOjEuNX0seyJpZCI6ImltcG9ydGVkLXRleHQtYmxvY2stMC1jb250ZW50LWVsZW1lbnQtNyIsInR5cGUiOiJ0ZXh0IiwieCI6MC4xMiwieSI6Mi45LCJsYWJlbCI6IlxcKEFcXCkiLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fSx7ImlkIjoiaW1wb3J0ZWQtdGV4dC1ibG9jay0wLWNvbnRlbnQtZWxlbWVudC04IiwidHlwZSI6InRleHQiLCJ4Ijo3LjE1LCJ5IjoxLjgsImxhYmVsIjoiXFwoQlxcKSIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC10ZXh0LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTkiLCJ0eXBlIjoidGV4dCIsIngiOjUuNzcsInkiOjYuNDIsImxhYmVsIjoiXFwoQ1xcKSIsImNvbG9yIjoiIzAwMDAwMCIsInN0cm9rZVdpZHRoIjoxLjV9LHsiaWQiOiJpbXBvcnRlZC10ZXh0LWJsb2NrLTAtY29udGVudC1lbGVtZW50LTEwIiwidHlwZSI6InRleHQiLCJ4IjoyLjU4LCJ5Ijo2LjgzLCJsYWJlbCI6IlxcKERcXCkiLCJjb2xvciI6IiMwMDAwMDAiLCJzdHJva2VXaWR0aCI6MS41fV0sImFsaWdubWVudCI6ImNlbnRlciJ9" 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 given diagram, $ABCD$ is a cyclic quadrilateral with $AD=BC$. Prove that&nbsp;$BD=AC$.</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/1768111145549-0sk3bi1rgxij.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTExNDU1NDktMHNrM2JpMXJneGlqLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI3MjU1NDAsImV4cCI6MTc5MjcyNTg0MH0.Y-loiciGCKdu5MdF2AXB4INlhtbOgNSrcdGWXz5PvdQ" style="max-width: min(170px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the given diagram, $ABCD$ is a cyclic quadrilateral with $AD=BC$. Prove that $BD=AC$.

<p>In the given diagram, $AP$ is tangent at $A$ to the smaller circle $AQB$, and $AQ$ is tangent at $A$ to the larger circle $APRB$ and $QBR$ is a straight line. Prove that&nbsp;$RP$ is parallel to $AQ$.</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/1768110489011-xeqgigjp67e.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA0ODkwMTEteGVxZ2lnanA2N2UucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.XVhqVTqcT3b_PlRb7g4_n-Qi7dPGOO285YHPWC-M_ww" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the given diagram, $AP$ is tangent at $A$ to the smaller circle $AQB$, and $AQ$ is tangent at $A$ to the larger circle $APRB$ and $QBR$ is a straight line. Prove that $RP$ is parallel to $AQ$.

<p>In the figure, $O$ is the centre of the circle $ABCD$. The smaller circle $AOKD$ intersects the larger circle at $A$ and $D$. Given that&nbsp;$AOB$, $AKC$ and $BKD$ 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">$DK = CK$,</div></li><li><div class="list-item-paragraph">$BKD$ bisects $\angle ODC$.</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/1768107139966-543ba3xiig6.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDcxMzk5NjYtNTQzYmEzeGlpZzYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.MOO0f71A1BMdyfv1SLxWPtLc1tBmsKfVZT3f8f-bCKs" style="max-width: min(170px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, $O$ is the centre of the circle $ABCD$. The smaller circle $AOKD$ intersects the larger circle at $A$ and $D$. Given that $AOB$, $AKC$ and $BKD$ 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzU1NjQzOTIzNDItcm55dDJza25zMGcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTU0MCwiZXhwIjoxNzkyNzI1ODQwfQ.18TeKZGSur5e71gTvSGXTOJi0HuQmCnoGHKZPMKA4tg" 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

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