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PSLE

PSLE Foundation Mathematics

<p>In the diagram, \(AB\) is a diameter of the circle with centre \(O\). \(DE\) and \(BF\) are tangents to the circle at \(C\) and \(B\) respectively. \(DCE\) and \(BEF\) are straight lines. Prove that</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1782643825419-7qc4gmhh4c5.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODI2NDM4MjU0MTktN3FjNGdtaGg0YzUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0MSwiZXhwIjoxNzkyNzI2MDQxfQ.pJ8JpctkJG5VkZt6H3rQqZw4Bfx-1wx5wFx6pcDQS7k" style="max-width: min(180px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-content"><span class="list-item-text">\(\Delta ABC\) and \(\Delta BFC\) are similar,</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-content"><span class="list-item-text">hence or otherwise, show that&nbsp;\(\Delta CFE\) is an isosceles triangle.</span><span class="inline-marks">[4]</span></div></li></ol>

In the diagram, \(AB\) is a diameter of the circle with centre \(O\). \(DE\) and \(BF\) are tangents to the circle at \(C\) and \(B\) respectively. \(DCE\) and \(BEF\) are straight lines. Prove that

\(\triangle ABC\) and \(\triangle BFC\) are similar,

hence or otherwise, show that \(\triangle CEF\) is an isosceles triangle.

Alternate Segment Theorem

Tangents Perpendicular to Radii

Similarity

Angles in a Semicircle and the Converse

Equal Tangents from an External Point

2025 AHS PRELIMS P1 Q13

Proofs in Plane Geometry

<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>
<div class="graph-block" data-graph-config="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" data-width="207" data-height="164" data-proportional style="width:207px;height:164px;max-width:100%;display:flow-root;margin:0 auto;--graph-width:207px;--graph-aspect:207 / 164;"></div>
<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 \(AC\).

Properties of Special Triangles

Angles in the Same Segment

Proving Concyclicity

Angles Formed by Two Parallel Lines and a Transversal

<p>In the diagram, \(TA\) and \(TB\) are tangents from \(T\) to the circle with centre \(O\) at \(A\) and \(B\), respectively. \(AC\) is a diameter of the circle. 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 \(AOB = 2 \times\) angle \(ACB\),</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(CB\) is parallel to \(OT\),</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">triangles \(OBC\) and \(TAB\) are similar, and</div></li><li data-list-item-id="item-iv" data-list-item-path="item-iv"><div class="list-item-paragraph">quadrilateral \(AOBT\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(A\), \(O\), \(B\) and \(T\).</div></li></ol>

In the diagram, \(TA\) and \(TB\) are tangents from \(T\) to the circle with centre \(O\) at \(A\) and \(B\), respectively. \(AC\) is a diameter of the circle. Prove that

triangles \(OBC\) and \(TAB\) are similar, and

quadrilateral \(AOBT\) is a cyclic quadrilateral, i.e., there exists a circle that passes through \(A\), \(O\), \(B\) and \(T\).

Angles at the Centre and Circumference

<p>In the diagram, \(SA\) is the tangent to the circle with centre \(O\) at \(A\) and \(AB\) is a diameter. \(C\) is a point on the circle and the tangent to the circle at \(C\) intersects \(SA\) at \(T\). \(BCS\) is a straight line. Prove that</p>
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<ol class="i-list"><li data-list-item-id="item-i" data-list-item-path="item-i"><div class="list-item-paragraph">angle \(CTS = 2 \times\) angle \(CAT\) ,</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(ST = CT\) , and</div></li><li data-list-item-id="item-iii" data-list-item-path="item-iii"><div class="list-item-paragraph">\(SC \times OC = AC \times AT\) .</div></li></ol>

In the diagram, \(SA\) is the tangent to the circle with centre \(O\) at \(A\) and \(AB\) is a diameter. \(C\) is a point on the circle and the tangent to the circle at \(C\) intersects \(SA\) at \(T\). \(BCS\) is a straight line. Prove that

<p>In the diagram, \(TA\) and \(TB\) are tangents to the circle from \(T\) at \(A\) and \(B\), respectively. Given that&nbsp;\(TA\) is parallel to \(BC\), 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">triangle \(ABC\) is an isosceles triangle, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">\(AB^2 = TB \times CB\).</div></li></ol>

In the diagram, \(TA\) and \(TB\) are tangents to the circle from \(T\) at \(A\) and \(B\), respectively. Given that \(TA\) is parallel to \(BC\), prove that

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

<p>In the diagram, the two circles touch each other internally at \(C\) and \(CD\) is the common tangent. \(AD\) and \(BD\) are tangents to the larger and smaller circles, 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 \(BCD\) and \(ACD\) are isosceles triangles, and</div></li><li data-list-item-id="item-ii" data-list-item-path="item-ii"><div class="list-item-paragraph">triangle \(ABD\) is an isosceles triangle.</div></li></ol>

In the diagram, the two circles touch each other internally at \(C\) and \(CD\) is the common tangent. \(AD\) and \(BD\) are tangents to the larger and smaller circles, respectively. Prove that

triangles \(BCD\) and \(ACD\) are isosceles triangles, and

triangle \(ABD\) is an isosceles triangle.

<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;"><p>In the figure, two circles touch at \(C\) and \(CD\) is a common tangent. \(AD\) and \(BD\) are tangents to the larger and smaller circles respectively. Show that&nbsp;\(AD=BD\).</p></div><div class="question-column" style="flex: 1; align-self: flex-start;"><div class="content-block"><figure class="question-figure alignright" style="text-align: right;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768111352957-lhmdq3pyno.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTEzNTI5NTctbGhtZHEzcHluby5wbmciLCJzY29wZSI6ImRvd25sb2FkIiwiaWF0IjoxNzkyNzI1NzQxLCJleHAiOjE3OTI3MjYwNDF9.w_WcnYLOKymoWbmDI8-1gMTYW-Y4hvKIyguQRk0X1zU" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

<p>In the figure, \(TA\) and \(TB\) are tangents to the circle, centre \(O\), from \(T\).</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;\(AOC\) is a diameter of the circle, state a reason why \(\angle AOB=2\angle ACB\).</div></li></ol><p>Hence, prove that</p><ol class="i-list" start="2"><li><div class="list-item-paragraph">\(CB\) is parallel to \(OT\),</div></li><li><div class="list-item-paragraph">A circle with diameter \(OT\) can be drawn to pass through \(A\) and \(B\),</div></li><li><div class="list-item-paragraph">\(\Delta OBC\) is similar to \(\Delta TAB\).</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/1768110672714-c5voub3uhv7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3NjgxMTA2NzI3MTQtYzV2b3ViM3VodjcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0MSwiZXhwIjoxNzkyNzI2MDQxfQ.pjPkgU0j5wxzlEx30noS8B97Zxee8Y57Jww8NvzeQio" style="max-width: min(200px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

In the figure, \(TA\) and \(TB\) are tangents to the circle, centre \(O\), from \(T\).

<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 style="text-align: left">The diagram shows a circle \(ABC\), with centre \(O\).<br>\(AP\) and \(BP\) are tangents to the circle.<br>\(\angle OAB = 25{}^\circ\)</p><ol class="abc-list"><li data-list-item-id="item-1780222062457-nj33mtt6n" data-list-item-path="item-1780222062457-nj33mtt6n"><div class="list-item-paragraph">Stating your reasons clearly, find</div><ol class="i-list"><li data-list-item-id="subitem-1780222066080-ekmhsk9a3" data-list-item-path="item-1780222062457-nj33mtt6n/subitem-1780222066080-ekmhsk9a3"><div class="list-item-paragraph">\(\angle AOB\).</div></li><li data-list-item-id="subitem-1780222066212-guqtj6m6s" data-list-item-path="item-1780222062457-nj33mtt6n/subitem-1780222066212-guqtj6m6s"><div class="list-item-paragraph">\(\angle PAB\).</div></li><li data-list-item-id="subitem-1780222066397-m8ji1u9yw" data-list-item-path="item-1780222062457-nj33mtt6n/subitem-1780222066397-m8ji1u9yw"><div class="list-item-paragraph">\(\angle APB\).</div></li></ol></li><li data-list-item-id="item-1780222063624-p3k2ioy91" data-list-item-path="item-1780222063624-p3k2ioy91"><div class="list-item-paragraph">For this part, use a separate configuration with the same tangency relationships, without the \(25^\circ\) condition from part (a). Given radius \(5\) cm and \(OP=16\) cm, find \(AP\).</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/1767602723698-5zgahl0huoa.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc2MDI3MjM2OTgtNXpnYWhsMGh1b2EucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0MSwiZXhwIjoxNzkyNzI2MDQxfQ.jyHIHMpG5RoBThX75RFH2N0OzL6RWdJvZ2B30TiZrB0" style="max-width: min(250px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

Applications of Pythagoras’ Theorem

Solving for Variables in Intersecting Circle Lines and Tangents

Properties of Circle

<p>\(TA\) and \(TB\) are tangents to a circle centre \(O\) and \(AT\) produced meets \(OB\) at \(Y\).<br>\(YB=12\) cm, \(TB=5\) cm and \(OA=r\) cm.</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">Write down the length of \(TA\).</div></li><li><div class="list-item-paragraph">Calculate the length of \(TY\).</div></li><li><div class="list-item-paragraph">Without solving <strong>(d)</strong>, find the numerical value of \(\frac{OA}{AY}\).</div></li><li><div class="list-item-paragraph">Calculate the value of \(r\).</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/1780571603336-2ndg69la94u.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3ODA1NzE2MDMzMzYtMm5kZzY5bGE5NHUucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0MSwiZXhwIjoxNzkyNzI2MDQxfQ.xFl3tTQvJ8rolfLlKdNBj3_XkYE7wxTI26-gkzQdhIg" style="max-width: min(350px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

\(TA\) and \(TB\) are tangents to a circle centre \(O\) and \(AT\) produced meets \(OB\) at \(Y\).
\(YB=12\) cm, \(TB=5\) cm and \(OA=r\) cm.

Calculation of Unknown Lengths in Circular Segments

<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, \(PA\) and \(PB\) are tangents to the circle with centre, \(O\).<br>\(PCO\) is a straight line where&nbsp;\(C\) is a point on the circle.<br>Find the</p><ol class="abc-list"><li><div class="list-item-paragraph">value of \(m\), where&nbsp;\(PC=m\) cm.</div></li><li><div class="list-item-paragraph">angle \(BPA\).</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/1767597257539-1jh37mgbrz8.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1OTcyNTc1MzktMWpoMzdtZ2JyejgucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0MSwiZXhwIjoxNzkyNzI2MDQxfQ.nO7ip3sRqGvBvvcx1yKtw6DKzd_hVKmXijmMb5xjVKU" style="max-width: min(280px, 100%); height: auto; display: block;" class="alignright custom-width" /></figure></div></div></div></div>

Basics of Trigonometric Ratios

Geometry Paper: Circle Theorems and Applications

Trigonometry

<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;"><p>The diagram shows a circle, centre \(O\) and radius \(3\) cm.<br>If&nbsp;\(CA\) and \(CB\) are tangents to the circle at \(A\) and \(B\) respectively and \(AC=8\) cm, find</p><ol class="abc-list"><li><div class="list-item-paragraph">the length of \(BC\),</div></li><li><div class="list-item-paragraph">the length of \(OC\),</div></li><li><div class="list-item-paragraph">the area of the quadrilateral \(AOBC\).</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/1767595729883-e98jjobh0d7.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njc1OTU3Mjk4ODMtZTk4ampvYmgwZDcucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjcyNTc0MSwiZXhwIjoxNzkyNzI2MDQxfQ.MyxndR4gWL_DAHtqg9X3F9EAjIaKRCHS2m7TE8iwzz0" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div></div></div></div>

Math Question Bank