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PSLE

PSLE Foundation Mathematics

<p>Triangle \(ABC\) is a right-angled triangle with angle \(ACB = 90^\circ\). \(ADEB\), \(BCGF\) and \(ACHI\) are squares. \(CJK\) is a straight line such that&nbsp;\(CJK\) is perpendicular to \(AB\) and \(DE\).<br>Show that</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">triangles \(DAC\) and \(BAI\) are congruent, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">the area of quadrilateral \(ADKJ\) and the area of square \(ACHI\) are equal;</span></div></li><li><div class="list-item-content"><span class="list-item-text">triangles \(BEC\) and \(BAF\) are congruent, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">the area of quadrilateral \(JKEB\) and the area of square \(BCGF\) are equal, hence</span></div></li><li><div class="list-item-content"><span class="list-item-text">\(AB^2 = AC^2 + BC^2\) (the Pythagoras Theorem).</span></div></li></ol>
<div class="content-block"><figure class="question-figure alignleft" style="text-align: left;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1770799522790-6382ys9wbpl.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTk1MjI3OTAtNjM4MnlzOXdicGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.g0AtMBXnXmFJptQhJDAkHEPiz_M1lxoPAVR223MbIh8" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignleft custom-width" /></figure></div>

Triangle \(ABC\) is a right-angled triangle with angle \(ACB = 90^\circ\). \(ADEB\), \(BCGF\) and \(ACHI\) are squares. \(CJK\) is a straight line such that \(CJK\) is perpendicular to \(AB\) and \(DE\).
Show that

triangles \(DAC\) and \(BAI\) are congruent, hence

the area of quadrilateral \(ADKJ\) and the area of square \(ACHI\) are equal;

triangles \(BEC\) and \(BAF\) are congruent, hence

the area of quadrilateral \(JKEB\) and the area of square \(BCGF\) are equal, hence

\(AB^2 = AC^2 + BC^2\) (the Pythagoras Theorem).

Area of Triangle

Area Ratios of Triangles with Equal Heights

Perimeter and Area of Rectangles and Squares

Congruence

IP2U8 HOT Q4 (Euclid's Proof of the Pythagoras' Theorem)

Congruence and Similarity

Perimeter and Area of Plane Figures

<p>In square \(ABCD\), \(E\) and \(F\) are points on sides \(CD\) and \(AD\), respectively, such that&nbsp;\(\angle AFB = 67^\circ\) and \(\angle CBE = 22^\circ\). Find \(\angle BEF\).</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/1770798524394-vlzlx63irqa.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTg1MjQzOTQtdmx6bHg2M2lycWEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.MZMmxBZAeowLFOupl_TNbjWN9JnGhFO3w4KpgE9eWUk" style="max-width: min(180px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In square \(ABCD\), \(E\) and \(F\) are points on sides \(CD\) and \(AD\), respectively, such that \(\angle AFB = 67^\circ\) and \(\angle CBE = 22^\circ\). Find \(\angle BEF\).

Geometric Transformations and Congruence

IP2U8 HOT Q2

<p>In triangle \(ABC\), \(\angle BAC = 45^\circ\), \(\angle ACB = 22.5^\circ\), \(BC = 12\) cm. Find the exact area of triangle \(ABC\).</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/1770798105175-922gq7pr9nq.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTgxMDUxNzUtOTIyZ3E3cHI5bnEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.yQxcKSzzZDR7moLC6OsD6eFuv_u9tOSY1z5ECqzpqh4" style="max-width: min(250px, 100%); height: auto; display: block;" alt="Triangle ABC with given angles and side length BC" class="alignright custom-width" /></figure></div>

In triangle \(ABC\), \(\angle BAC = 45^\circ\), \(\angle ACB = 22.5^\circ\), \(BC = 12\) cm. Find the exact area of triangle \(ABC\).

IP2U8 HOT Q1

<p>In the diagram below, \(\Delta ADG\) and \(\Delta EHC\) overlap to form a parallelogram \(BDFH\).</p>
<p>Given further that&nbsp;\(AG = EC\) and \(\angle HCE = \angle DGA\), prove that&nbsp;\(\Delta ADG\) is congruent to \(\Delta EHC\). State your reasons clearly.</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/1770792023650-kua5hxx2a0l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTIwMjM2NTAta3VhNWh4eDJhMGwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.pGS98ASAQ_6TF0USjLM8r1TB3lduTnDshIopDe6xXuI" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram below, \(\Delta ADG\) and \(\Delta EHC\) overlap to form a parallelogram \(BDFH\).

Given further that \(AG = EC\) and \(\angle HCE = \angle DGA\), prove that \(\Delta ADG\) is congruent to \(\Delta EHC\). State your reasons clearly.

IP2U8 Exercise Q4

<p>In the triangle below, \(C\) and \(D\) are on the line \(BE\) such that&nbsp;\(AB = AE\) and \(AC = AD\).</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>Prove that triangle \(ABC\) is congruent to triangle \(AED\).</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/1770791513482-26doz19xh4.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTE1MTM0ODItMjZkb3oxOXhoNC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzU4LCJleHAiOjE3OTI3MjYwNTh9._igXddUqx1lrZFefC_3Fg5joZSidvr7XZZzZz0zP4eE" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the triangle below, \(C\) and \(D\) are on the line \(BE\) such that \(AB = AE\) and \(AC = AD\).

IP2U8 Exercise Q3

<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, the straight line \(ACD\) bisects \(\angle ECB\).<br>\(BC = CE\) and \(AD\) is perpendicular to \(EB\).<br></p><ol class="i-list"><li><div class="list-item-paragraph">Name a pair of congruent triangles.</div></li><li><div class="list-item-paragraph">Give reasons for the congruence.</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/1770790824149-2ttsw1pacuf.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTA4MjQxNDktMnR0c3cxcGFjdWYucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.RHVJ0k17yiDlHuvTp5SdVQhpHVQ8R0lwstSR2Tzezf0" style="max-width: min(280px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

IP2U8 Exercise Q2

<p>In the diagram, \(ABCD\) and \(DEFG\) are squares on a flat surface.</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;\(\angle ADE = \angle CDG\).</div></li><li><div class="list-item-paragraph">Prove that&nbsp;\(\Delta ADE\) is congruent to \(\Delta CDG\).</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/1770790267425-62guh5icqje.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3OTAyNjc0MjUtNjJndWg1aWNxamUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.FqUMC6M0V3wSx9o5kfiOiS9CJwln0TsyRzUN97giGMs" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, \(ABCD\) and \(DEFG\) are squares on a flat surface.

IP2U8 Exercise Q1

<p>In the diagram, \(P\) is a point on the inside of the square \(ABCD\) such that&nbsp;\(PA = 2\) cm, \(PB = 5\) cm and \(PD = \sqrt{17}\) cm. Find the exact area of square \(ABCD\).</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/1770789331534-kjh2q0njnp.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3ODkzMzE1MzQta2poMnEwbmpucC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzU4LCJleHAiOjE3OTI3MjYwNTh9.CyB9DI1Q_CqBagNaZ4kZbK8gxWhVyAsW0IaZ2_5vLiQ" style="max-width: min(150px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the diagram, \(P\) is a point on the inside of the square \(ABCD\) such that \(PA = 2\) cm, \(PB = 5\) cm and \(PD = \sqrt{17}\) cm. Find the exact area of square \(ABCD\).

Applications of Pythagoras’ Theorem

The Converse of Pythagoras Theorem

IP2U7 HOT Q3

Pythagoras’ Theorem

<p>In the diagram, \(O\) is a point on the inside of the equilateral triangle \(ABC\) such that&nbsp;\(OA = 3\) cm, \(OB = 4\) cm and \(OC = 5\) cm. Find \(\angle AOB\).</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/1770789024351-xi9mys6qi3.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3ODkwMjQzNTEteGk5bXlzNnFpMy5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzU4LCJleHAiOjE3OTI3MjYwNTh9.CZmD7ik9fs79Btul3SlCnOM3N3uECN4Eq0WPPgTV2S0" style="max-width: min(170px, 100%); height: auto; display: block;" alt="Equilateral triangle ABC with internal point O" class="alignright custom-width" /></figure></div>

In the diagram, \(O\) is a point on the inside of the equilateral triangle \(ABC\) such that \(OA = 3\) cm, \(OB = 4\) cm and \(OC = 5\) cm. Find \(\angle AOB\).

IP2U7 HOT 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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NzA3MjQ5NzA0NTQtNjBqN3c2a2RrcWUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.0BUtaF7wfIsHgmHvSeVYmuRq0XYbSeiRttnrYPcFD-o" 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 \(P\).

Angles in the Same Segment

IP2U14 Exercise Q1

Properties of Circle

Proofs in Plane Geometry

<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

Alternate Segment Theorem

Angles in Cyclic Quadrilaterals

Similarity

<p>In the diagram, \(ABCD\) is a cyclic quadrilateral with \(AD = BC\). 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">\(BD = AC\), 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

Angles Formed by Two Parallel Lines and a Transversal

<p>In parallelogram \(ABCD\), \(E\), \(F\) are on diagonal \(AC\) such that&nbsp;\(AE = CF\). \(BEH\) and \(DFG\) 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">triangles \(BAE\) and \(DCF\) are congruent,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles \(BAH\) and \(DCG\) are congruent, and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">quadrilateral \(BGDH\) is a parallelogram.</div></li></ol>

In parallelogram \(ABCD\), \(E\), \(F\) are on diagonal \(AC\) such that \(AE = CF\). \(BEH\) and \(DFG\) are straight lines. Prove that

triangles \(BAE\) and \(DCF\) are congruent,

triangles \(BAH\) and \(DCG\) are congruent, and

quadrilateral \(BGDH\) is a parallelogram.

<p>In the diagram, \(ABF\), \(CAG\) and \(EFG\) are straight lines. Given that&nbsp;\(AG = AC = CF\) and \(AF = FB\), 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 \(AFG\) and \(FBC\) are congruent, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle \(BFE\) is an isosceles triangle.</div></li></ol>

In the diagram, \(ABF\), \(CAG\) and \(EFG\) are straight lines. Given that \(AG = AC = CF\) and \(AF = FB\), prove that

triangles \(AFG\) and \(FBC\) are congruent, and

triangle \(BFE\) is an isosceles triangle.

Properties of Special Triangles

<p>In the diagram, angle \(BAC\) and angle \(ABD\) are right angles. \(AB = AC\), \(AF = \frac{2}{3} AB\) and \(BD = \frac{2}{3} AC\). \(AD\), \(CF\) intersect at \(E\). 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 \(ACF\) and \(BAD\) are congruent, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangles \(ACF\) and \(EAF\) are similar.</div></li></ol>

In the diagram, angle \(BAC\) and angle \(ABD\) are right angles. \(AB = AC\), \(AF = \frac{2}{3} AB\) and \(BD = \frac{2}{3} AC\). \(AD\), \(CF\) intersect at \(E\). Prove that

triangles \(ACF\) and \(BAD\) are congruent, and

triangles \(ACF\) and \(EAF\) are similar.

<p>In the diagram, \(ABCD\) and \(DEFG\) are squares. Prove that</p>
<div class="graph-block" data-graph-config="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" data-width="222" data-height="301" data-proportional style="width:222px;height:301px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:222px;--graph-aspect:222 / 301;"></div>
<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">triangles \(ADE\) and \(CDG\) are congruent, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(AE\) is perpendicular to \(CG\).</div></li></ol>

In the diagram, \(ABCD\) and \(DEFG\) are squares. Prove that

triangles \(ADE\) and \(CDG\) are congruent, and

<p>In trapezium \(ABGD\), \(AB\) is parallel to \(DC\) and \(E\) and \(F\) are the midpoints of the line segments \(BD\) and \(AC\) respectively. \(BFG\) and \(DGC\) 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-1775568004152-b1gem4j5w" data-list-item-path="item-1775568004152-b1gem4j5w"><div class="list-item-paragraph">\(\Delta ABF\) and \(\Delta CGF\) are congruent,</div></li><li data-list-item-id="item-1775568004660-m22p158mh" data-list-item-path="item-1775568004660-m22p158mh"><div class="list-item-paragraph">\(EF=\frac{1}{2}(CD-AB)\).</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/1768111956418-6veigrna3ew.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE5NTY0MTgtNnZlaWdybmEzZXcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.qECaJxWmR4qqFC_m8r37h39rZ_rIOKmwqRD2yArL8wY" style="max-width: min(250px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In trapezium \(ABGD\), \(AB\) is parallel to \(DC\) and \(E\) and \(F\) are the midpoints of the line segments \(BD\) and \(AC\) respectively. \(BFG\) and \(DGC\) are straight lines. Prove that

Midpoint Theorem

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTExNDU1NDktMHNrM2JpMXJneGlqLnBuZyIsInNjb3BlIjoiZG93bmxvYWQiLCJpYXQiOjE3OTI3MjU3NTgsImV4cCI6MTc5MjcyNjA1OH0.QF2TLE8SGBiTn0tZJqPlAD74Q3j2j7ZJnTLuRqmfclI" 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 trapezium \(ABCD\), \(AB // DC\) and \(P\) and \(Q\) are the midpoints of sides \(AD\) and \(BC\) respectively. The extensions of \(BP\) and \(CD\) intersect at \(S\). 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-1780397797725-q7qhkxok7"><div class="list-item-paragraph">\(P\) is the midpoint of \(BS\), hence show that&nbsp;\(PQ // DC\),</div></li><li data-list-item-id="item-1780397798229-2fa2yc9je"><div class="list-item-paragraph">\(PQ = \frac{1}{2}(AB + DC)\).</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/1768108542681-izhn480r1ns.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg1NDI2ODEtaXpobjQ4MHIxbnMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc1OCwiZXhwIjoxNzkyNzI2MDU4fQ.P_dRfdC-GjtYJLXWUXQNE3D0YMpWGAtIYluHzWN7tLI" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the trapezium \(ABCD\), \(AB // DC\) and \(P\) and \(Q\) are the midpoints of sides \(AD\) and \(BC\) respectively. The extensions of \(BP\) and \(CD\) intersect at \(S\). Prove that

<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, \(BE=AD\), \(AE=BD\) and \(\angle ACE=76{}^\circ\).</p><ol class="i-list"><li><div class="list-item-paragraph">Show that&nbsp;\(\Delta ABE\) and \(\Delta BAD\) are congruent.</div></li><li><div class="list-item-paragraph">Find \(\angle ABE\).</div></li><li><div class="list-item-paragraph">Name another pair of congruent triangles.</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/1767494373228-s8yvw6nq2p.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc0OTQzNzMyMjgtczh5dnc2bnEycC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzU4LCJleHAiOjE3OTI3MjYwNTh9.n6WvSgWomwM0Mf19jZk6jcfzX4R8I3u2Xy8BWdEqdMw" style="max-width: min(320px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

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