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PSLE

PSLE Foundation Mathematics

<p>In triangle \(ABC\), \(M\) and \(N\) are the midpoints of \(AB\) and \(AC\), respectively. \(PMNQ\) is a straight line such that&nbsp;\(BP\), \(CQ\) are perpendicular to \(PMNQ\). 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">quadrilateral \(BCQP\) is a rectangle, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">area of triangle \(ABC =\) area of quadrilateral \(BCQP\).</div></li></ol>

In triangle \(ABC\), \(M\) and \(N\) are the midpoints of \(AB\) and \(AC\), respectively. \(PMNQ\) is a straight line such that \(BP\), \(CQ\) are perpendicular to \(PMNQ\). Prove that

quadrilateral \(BCQP\) is a rectangle, and

area of triangle \(ABC =\) area of quadrilateral \(BCQP\).

Properties of Special Quadrilaterals

Midpoint Theorem

Proofs in Plane Geometry

<p>In the rhombus \(ABCD\), \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(AD\), 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">quadrilateral \(PQRS\) is a rectangle.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Generalise the above result by stating one sufficient condition that quadrilateral \(ABCD\) must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of \(ABCD\) is a rectangle.</div></li></ol>

In the rhombus \(ABCD\), \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(AD\), respectively. Prove that

Generalise the above result by stating one sufficient condition that quadrilateral \(ABCD\) must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of \(ABCD\) is a rectangle.

<p>In rectangle \(ABCD\), \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(AD\), 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">quadrilateral \(PQRS\) is a rhombus.</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">Generalise the above result by stating one sufficient condition that quadrilateral \(ABCD\) must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of \(ABCD\) is a rhombus.</div></li></ol>

In rectangle \(ABCD\), \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(AD\), respectively. Prove that

Generalise the above result by stating one sufficient condition that quadrilateral \(ABCD\) must satisfy, such that the quadrilateral formed by joining the midpoints of the four sides of \(ABCD\) is a rhombus.

<p>In quadrilateral \(ABCD\), \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(AD\), respectively. Prove that quadrilateral \(PQRS\) is a parallelogram.</p>
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In quadrilateral \(ABCD\), \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(BC\), \(CD\) and \(AD\), respectively. Prove that quadrilateral \(PQRS\) is a parallelogram.

<p>In triangle \(ABC\), \(BQ\) and \(CP\) intersect at \(O\). \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(AC\), \(OC\) and \(OB\) 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">\(PQRS\) is a parallelogram,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(OS = \frac{1}{3} BQ\), and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">\(BS = \frac{1}{3} BQ\).</div></li></ol>

In triangle \(ABC\), \(BQ\) and \(CP\) intersect at \(O\). \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(AC\), \(OC\) and \(OB\) respectively. Prove that

<p>In the diagram, \(AB\) is parallel to \(PRQ\), \(T\) and \(P\) are the midpoints of \(CQ\) and \(CA\), respectively, and \(ART\) is a straight line. Prove that&nbsp;\(AR = 2RT\).</p>
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In the diagram, \(AB\) is parallel to \(PRQ\), \(T\) and \(P\) are the midpoints of \(CQ\) and \(CA\), respectively, and \(ART\) is a straight line. Prove that \(AR = 2RT\).

Similarity

<p>In the diagram, \(BFD\), \(AFC\) and \(BEC\) are straight lines. It is given that&nbsp;\(AF = FC\), \(BF = FD\) and \(BE = EC\). 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">\(FE = \frac{1}{2} AB\), and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(AB = DC\).</div></li></ol>

In the diagram, \(BFD\), \(AFC\) and \(BEC\) are straight lines. It is given that \(AF = FC\), \(BF = FD\) and \(BE = EC\). Prove that

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE5NTY0MTgtNnZlaWdybmEzZXcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.glQdcuUgR4KRUTle4_LmSHxFbUoEcBwS_2Exm3QqKx0" 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

Congruence

<p>In \(\Delta ABC\), \(E\) is the midpoint of \(AC\). \(AD\) and \(BE\) intersect at \(F\), which is the midpoint of \(BE\). \(FC\) and \(EH\) intersect at \(G\).</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2-1" data-layout="2-1"><div class="question-column" style="flex: 2; align-self: flex-start;"><ol class="i-list"><li><div class="list-item-paragraph">Given that&nbsp;\(AD\) is parallel to \(EH\), prove that&nbsp;\(EH=2FD\).</div></li><li><div class="list-item-paragraph">Hence, show that&nbsp;\(AF=3FD\).</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/1768111671039-bfve9sm655g.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTE2NzEwMzktYmZ2ZTlzbTY1NWcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.b711GC9Qd2Zs4OorguePyggV72HfMTq1XloN_snk6is" style="max-width: min(300px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In \(\Delta ABC\), \(E\) is the midpoint of \(AC\). \(AD\) and \(BE\) intersect at \(F\), which is the midpoint of \(BE\). \(FC\) and \(EH\) intersect at \(G\).

<p>In the diagram, \(E\), \(F\), \(G\) and \(H\) are the midpoints of the sides of the rectangle \(ABCD\). The points \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(HE\), \(EF\), \(FG\) and \(GH\) respectively. By drawing the line segments \(EG\) and \(HF\), 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><div class="list-item-paragraph">\(EF = EH\),</div></li><li><div class="list-item-paragraph">\(EFGH\) is a rhombus,</div></li><li><div class="list-item-paragraph">\(\angle SPQ = 90{}^\circ\),</div></li><li><div class="list-item-paragraph">\(PQRS\) is a rectangle.</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/1768109085978-hw9eg7wq788.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDkwODU5NzgtaHc5ZWc3d3E3ODgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.S0B4aHiPQCQ0kiCwTciciiJU0Mg7RzZOI2qIR4dHZmA" style="max-width: min(240px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, \(E\), \(F\), \(G\) and \(H\) are the midpoints of the sides of the rectangle \(ABCD\). The points \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(HE\), \(EF\), \(FG\) and \(GH\) respectively. By drawing the line segments \(EG\) and \(HF\), prove that

Applications of Pythagoras’ Theorem

<p>Quadrilateral \(QUAD\) has an inscribed quadrilateral \(PROF\), where&nbsp;\(P\), \(R\), \(O\) and \(F\) are the midpoints of \(QD\), \(QU\), \(UA\) and \(AD\) respectively. Prove that quadrilateral \(PROF\) is a parallelogram.</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/1768109018332-puqlk12l57f.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDkwMTgzMzItcHVxbGsxMmw1N2YucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.eRO4QoLkZpVkz6DYbk-LTMeBxk59VvAB249lvRJBSNk" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

Quadrilateral \(QUAD\) has an inscribed quadrilateral \(PROF\), where \(P\), \(R\), \(O\) and \(F\) are the midpoints of \(QD\), \(QU\), \(UA\) and \(AD\) respectively. Prove that quadrilateral \(PROF\) is a parallelogram.

<p>In \(\Delta ABC\), \(BQ\) and \(CP\) intersect at \(O\). \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(AC\), \(OC\) and \(OB\) respectively. Prove that&nbsp;\(PQRS\) is a parallelogram. Hence, show that&nbsp;\(OS = \frac{1}{3} BQ\).</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/1768108924862-6o0dkhqg89l.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg5MjQ4NjItNm8wZGtocWc4OWwucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.-vQQqSbeRAHU2JlXq9NDDxsdIDNqhDNGFwJs9CK9ZIM" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In \(\Delta ABC\), \(BQ\) and \(CP\) intersect at \(O\). \(P\), \(Q\), \(R\) and \(S\) are the midpoints of \(AB\), \(AC\), \(OC\) and \(OB\) respectively. Prove that \(PQRS\) is a parallelogram. Hence, show that \(OS = \frac{1}{3} BQ\).

Proving Quadrilaterals

<p>In the diagram, \(AD = DC\), \(AE = EB\), and the lines \(BD\) and \(CE\) of \(\Delta ABC\) intersect at \(G\). The points \(F\) and \(H\) are the midpoints of \(BG\) and \(CG\) respectively. 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><div class="list-item-paragraph">\(DEFH\) is a parallelogram,</div></li><li><div class="list-item-paragraph">\(BF = DG\) and \(CH = GE\).</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/1768108734383-v2cbjs5694.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg3MzQzODMtdjJjYmpzNTY5NC5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NjMyLCJleHAiOjE3OTI3MjU5MzJ9.S0ajkEGU2Nl47ONfF1Jl0G7Jta4igzw4NCDRoAEZXd8" style="max-width: min(200px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

In the diagram, \(AD = DC\), \(AE = EB\), and the lines \(BD\) and \(CE\) of \(\Delta ABC\) intersect at \(G\). The points \(F\) and \(H\) are the midpoints of \(BG\) and \(CG\) respectively. Prove that

<p>In the diagram, \(EFG\) is a straight line. The points \(E\), \(F\) and \(G\) are the midpoints of \(AD\), \(AC\) and \(BC\) respectively. 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><div class="list-item-paragraph">\(EG = \frac{1}{2}(AB + DC)\),</div></li><li><div class="list-item-paragraph">\(\angle CAB = \angle ACD\).</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/1768108652165-vzdx6az529k.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg2NTIxNjUtdnpkeDZhejUyOWsucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.06QxaA4MPsqBvkISxGksAUsbOsjtR4rHtpBZTvn48Z4" style="max-width: min(180px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, \(EFG\) is a straight line. The points \(E\), \(F\) and \(G\) are the midpoints of \(AD\), \(AC\) and \(BC\) respectively. Prove that

Angles Formed by Two Parallel Lines and a Transversal

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg1NDI2ODEtaXpobjQ4MHIxbnMucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.OetvU4D-AKBIxSQaKM2n0ttLuV_fOsZs3MvL_OyqjsI" 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

<p>In the figure, \(AB\) is parallel to \(PRQ\), \(T\) and \(P\) are the midpoints of \(CQ\) and \(CA\) respectively and \(ART\) is a straight line. Prove that&nbsp;\(TR : RA = 1 : 2\).</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/1768108469612-lbo2ru8jmsr.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDg0Njk2MTItbGJvMnJ1OGptc3IucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTYzMiwiZXhwIjoxNzkyNzI1OTMyfQ.sAfC_tU6mgFySabZqXAtuOy2BGxyr1mV3PNL6RvDFHg" style="max-width: min(220px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div>

In the figure, \(AB\) is parallel to \(PRQ\), \(T\) and \(P\) are the midpoints of \(CQ\) and \(CA\) respectively and \(ART\) is a straight line. Prove that \(TR : RA = 1 : 2\).

<p>In the diagram, \(BFD\), \(AFC\) and \(BEC\) are straight lines. It is given that&nbsp;\(AF = FC\), \(BF = FD\), \(BE = EC\). 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">\(FE = \frac{1}{2} AB\),</div></li><li><div class="list-item-paragraph">\(AB = DC\).</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/1768108373506-y07s2ef80e.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMDgzNzM1MDYteTA3czJlZjgwZS5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NjMyLCJleHAiOjE3OTI3MjU5MzJ9.u_W3ZQWl6dkiH3keYjPDlVkB678yCP3mVFOwjH2eTdk" style="max-width: min(230px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the diagram, \(BFD\), \(AFC\) and \(BEC\) are straight lines. It is given that \(AF = FC\), \(BF = FD\), \(BE = EC\). Prove that

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