A Math Plane Geometry Practices

Page 1 Paradigm Specialising in O Level Mathematics Maths Secrets Plane Geometry 1 In the figure, 𝑋𝑌𝑍 is a straight line that is tangent to the circle at 𝑋. 𝑋𝑄 bisects ∠𝑅𝑋𝑍 and cuts the circle at 𝑆. 𝑅𝑆 produced meets 𝑋𝑍 at 𝑌 and 𝑍𝑅 = 𝑋𝑅. Prove that a) 𝑆𝑅 = 𝑆𝑋, b) a circle can be drawn passing through 𝑍, 𝑌, 𝑆 and 𝑄. 2 In the diagram, 𝐴, 𝐵, 𝐶 and 𝐷 are points on the circle centre 𝑂. 𝐴𝑃 and 𝐵𝑃 are tangents to the circle at 𝐴 and 𝐵 respectively. 𝐷𝑄 and 𝐶𝑄 are tangents to the circle at 𝐷 and 𝐶 respectively. 𝑃𝑂𝑄 is a straight line. (i)Prove that angle 𝐶𝑃𝐷 = 2 × angle 𝐶𝐷𝑄. (ii)Make a similar deduction about angle 𝐴𝑂𝐵. (iii)Prove that 2 × angle 𝑂𝐴𝐷 = angle 𝐶𝐷𝑄 + angle 𝐵𝐴𝑃 3 The diagram shows two intersecting circles, 𝐶1 and 𝐶2. 𝐶1 passes through the vertices of the triangle 𝐴𝐵𝐷. The tangents to 𝐶1 at 𝐴 and 𝐵 intersect at the point 𝑄 on 𝐶2. A line os drawn from 𝑄 to intersect the line 𝐴𝐷 at 𝐸 on 𝐶2. Prove that (i)𝑄𝐸 bisects angle 𝐴𝐸𝐵, (ii)𝐸𝐵 = 𝐸𝐷, (iii)𝐵𝐷 is parallel to 𝑄𝐸. 4 In the diagram, 𝐴, 𝐵 and 𝐶 are three points on the circle such that 𝐴𝐵 is the diameter of the circle and 𝑊 is the midpoint of 𝐴𝐶. 𝐴𝐵 and 𝐶𝐾 are parallel to each other and 𝐾𝐿 is a tangent to the circle at 𝐴 (i)Prove that 𝑂𝑊 is parallel to 𝐵𝐶. (ii)Prove that Angle 𝐴𝑊𝑂 = Angle 𝐴𝐾𝐶.

Page 2 Paradigm Specialising in O Level Mathematics Maths Secrets 5 The diagram shows a point 𝑃 on a circle and 𝑃𝑄 is a tangent to the circle. Points 𝐴, 𝐵 and 𝐶 lie on the circle such that 𝑃𝐴 bisects angle 𝑄𝑃𝐵 and 𝑄𝐴𝐶 is a straight line. The lines 𝑄𝐶 and 𝑃𝐵 intersect at 𝐷. (i) Prove that 𝐴𝑃 = 𝐴𝐵. (ii) Prove that 𝐶𝐷 bisects angle 𝑃𝐶𝐵. (iii) Prove that triangles 𝐶𝐷𝑃 and 𝐶𝐵𝐴 are similar. 6 The diagram shows a circle passing through points 𝐷, 𝐸, 𝐶 and 𝐹, where 𝐹𝐶 = 𝐹𝐷. The point 𝐷 lies on 𝐴𝑃 such that 𝐴𝐷 = 𝐷𝑃. 𝐷𝐶 and 𝐸𝐹 cut 𝑃𝐵 at 𝑇 such that 𝑃𝑇 = 𝑇𝐵. (i) Show that 𝐴𝐵 is a tangent to the circle at point 𝐹. (ii) By showing that triangle 𝐷𝐹𝑇 and triangle 𝐸𝐹𝐷 are similar show that 𝐷𝐹2 − 𝐹𝑇2 = 𝐹𝑇 × 𝐸𝑇. 7 Given that 𝐴𝐷 and 𝐵𝐶 are straight lines, 𝐴𝐶 bisects angle 𝐷𝐴𝑌 and 𝐴𝐵 bisects angle 𝐷𝐴𝑋, show that (i) 𝐴𝐶2 = 𝐸𝐶 × 𝐵𝐶, (ii) 𝐵𝐶 is a diameter of the circle, (iii) 𝐴𝐷 and 𝐵𝐶 are perpendicular to each other. 8 In the diagram, two circles touch each other at 𝐴. 𝑇𝐴 is tangent to both circles at 𝐴 and 𝐹𝐸 is a tangent to the smaller circle at 𝐶. Chords 𝐴𝐸 and 𝐴𝐹 intersect the smaller circle at 𝐵 and 𝐷 respectively. Prove that (i) line 𝐵𝐷 is parallel to line 𝐹𝐸, (ii) ∠𝐹𝐴𝐶 = ∠𝐶𝐴𝐸. 9 In the diagram, 𝐴𝐶𝐷𝐸 is a cyclic quadrilateral. Lines 𝐺𝐴𝐵 and 𝐹𝐸𝐻𝐶 are parallel, and line 𝐺𝐴𝐵 is a tangent to the circle at 𝐴. Lines 𝐴𝐷 and 𝐸𝐶 meet at 𝐻. Prove that (i) triangle 𝐴𝐵𝐷 and triangle 𝐶𝐵𝐴 are similar, (ii) triangle 𝐴𝐶𝐻 and triangle 𝐴𝐷𝐶 are similar, (iii) 𝐴𝐷 bisects angle 𝐶𝐷𝐸, (iv) 𝐴𝐵 × 𝐴𝐻 = 𝐴𝐶 × 𝐵𝐶.

Page 3 Paradigm Specialising in O Level Mathematics Maths Secrets 10 The diagram shows two circles that intersect each other at points 𝐴 and 𝐶. The points 𝐸 and 𝐷 lie on the circumference of the larger circle. The point 𝐵 lies on the circumference of the smaller circle such that 𝐵𝐶𝐷 is a straight line. Line 𝐶𝐹 is a tangent to the smaller circle at 𝐶. 𝐴𝐶 = 𝐵𝐶 and 𝐴𝐸 = 𝐸𝐷. (i) Prove that 𝐴𝐵 and 𝐶𝐹 are parallel. (ii) Prove that ∆𝐴𝐵𝐶 is similar to ∆𝐴𝐷𝐸 and hence show that 𝐴𝐵 × 𝐷𝐸 = 𝐴𝐷 × 𝐵𝐶.

Page 4 Paradigm Specialising in O Level Mathematics Maths Secrets Answers 1 (a) ∠𝑍𝑋𝑄 = ∠𝑆𝑅𝑋 (Alternate Segment Theorem) ∠𝑍𝑋𝑄 = ∠𝑄𝑋𝑅 (XQ is the angle bisector of ∠𝑅𝑋𝑍) ∠𝑄𝑋𝑅 = ∠𝑆𝑅𝑋 By base angles of isosceles triangles, 𝑆𝑅 = 𝑆𝑋 (b) Let ∠𝑄𝑋𝑅 be x ∠𝑅𝑆𝑋 = 180˚ − 2𝑥 (Isosceles Triangle) ∠𝑌𝑆𝑄 = 180˚ − 2𝑥 (Vertically Opposite Angles) ∠𝑅𝑍𝑋 = ∠𝑍𝑋𝑅 = 2𝑥 (Base angles of Isosceles Triangle) ∠𝑅𝑍𝑋 + ∠𝑌𝑆𝑄 = 180˚ − 2𝑥 + 2𝑥 = 180˚ Since opposite angles are supplementary in cyclic quadrilaterals, a circle that passes through Z, Y, S and Q can be drawn. Alternative Similar but use of tangent secant theorem. 2 Let ∠𝐶𝐷𝑄 = 𝑎 ∠𝑂𝐷𝑄 = 90˚ (tan ┴ rad) ∴∠𝑂𝐷𝐶 = 90˚ − 𝑎 ∴∠𝐶𝑂𝐷 = 180˚ − 2(90˚ − 𝑎)(∠sum, ∆𝐶𝑂𝐷) ∠𝐴𝑂𝐵 = 2 × ∠𝐵𝐴𝑃 From (i) and (ii), 2(∠𝐶𝐷𝑄 + ∠𝐵𝐴𝑃 = ∠𝐶𝑂𝐷 + ∠𝐴𝑂𝐵 ∠𝐶𝐷𝑄 + ∠𝐵𝐴𝑃 = 1 2 (∠𝐶𝑂𝐷 + ∠𝐴𝑂𝐵) = ∠𝐴𝑂𝑃 + ∠𝐷𝑂𝑄 (┴ prop of chord) = 180˚ − ∠𝐴𝑂𝐷 = 2∠𝑂𝐴𝐷 3 (i) Let ∠𝑄𝐸𝐴 = 𝑥˚ ∠𝑄𝐵𝐴 = ∠𝑄𝐸𝐴 (angles in same segment in C2) B1 = 𝑥˚ 𝑄𝐵 = 𝑄𝐴 (tangents to C1 from external point Q) B1 ∠𝑄𝐴𝐵 = ∠𝑄𝐵𝐴 (base angles of isosceles triangle) B1 = 𝑥˚ ∴ ∠𝑄𝐸𝐵 = ∠𝑄𝐸𝐴 Hence, QE bisects angle AEB. (ii) ∠𝑄𝐵𝐴 = 𝑥˚ (from (i)) ∠𝐴𝐷𝐵 = ∠𝑄𝐵𝐴 (angles in alternate segment in C1) either = 𝑥˚ ∠𝐴𝐸𝐵 = 2𝑥˚ (from (i)) ∠𝐷𝐵𝐸 = ∠𝐴𝐸𝐵 − ∠𝐴𝐷𝐵 (exterior angle of triangle BDE) or B1 = 2𝑥˚ − 𝑥˚ = 𝑥˚ ∴∠𝐴𝐷𝐵 = ∠𝐸𝐷𝐵 = ∠𝐷𝐵𝐸 = 𝑥˚ (base angles of isosceles triangle BDE) B1 Hence 𝐸𝐵 = 𝐸𝐷 (iii) [2] From (i) ∠𝐸𝐵𝐷 = ∠𝑄𝐸𝐵 = 𝑥 B1 ∴∠𝐸𝐵𝐷 and ∠𝑄𝐸𝐵 are alternate angles of parallel lines. (alternate angles are equal) B1 BD is parallel to QE

Page 5 Paradigm Specialising in O Level Mathematics Maths Secrets 4 O is the midpoint of AB and W is the midpoint of AC. By Midpoint Theorem, BC is parallel to OW. Angle 𝐴𝑂𝑊 = Angle 𝐴𝐵𝐶 (corr angles, OW||BC) Angle 𝐴𝐵𝐶 = Angle 𝐶𝐴𝐾 (alt segment theorem) → 𝐴𝑛𝑔𝑙𝑒 𝐴𝑂𝑊 = 𝐴𝑛𝑔𝑙𝑒 𝐶𝐴𝐾 Angle 𝐵𝐴𝐶 = 𝐴𝑛𝑔𝑙𝑒 𝐴𝐶𝐾 (𝑎𝑙𝑡 𝑎𝑛𝑔𝑙𝑒𝑠, 𝐴𝐵||CK) ∴Angle AWO = 180˚ − 𝐴𝑛𝑔𝑙𝑒 𝐵𝐴𝐶 − 𝐴𝑛𝑔𝑙𝑒 𝐴𝑂𝑊 (𝐴𝑛𝑔𝑙𝑒 𝑠𝑢𝑚 𝑜𝑓 △) = 180˚ − 𝐴𝑛𝑔𝑙𝑒 𝐴𝐶𝐾 − 𝐴𝑛𝑔𝑙𝑒 𝐶𝐴𝐾 = 𝐴𝑛𝑔𝑙𝑒 𝐴𝐾𝐶 (shown) 5 (i) ∠𝐴𝐵𝑃 = ∠𝐴𝑃𝑄 (alt. segment theorem) Since PA bisects ∠𝑄𝑃𝐵, ∠𝐴𝑃𝑄 = ∠𝐴𝑃𝐵 ∴∠𝐴𝐵𝑃 = ∠𝐴𝑃𝐵 (base ∠𝑠 of isosceles triangle APB) Hence, 𝐴𝑃 = 𝐴𝐵. (i) ∠𝐴𝐶𝐵 = ∠𝐴𝑃𝐵 (∠𝑠 in the same segment) ∠𝐴𝐶𝑃 = ∠𝐴𝐵𝑃 (∠𝑠 in the same segment) = ∠𝐴𝑃𝐵 (shown) ∠𝐴𝐶𝐵 = ∠𝐴𝐶𝑃 Hence, CD bisects ∠𝑃𝐶𝐵. (ii) ∠𝐴𝐶𝐵 = ∠𝐴𝐶𝑃 (from ii) ∠𝐶𝑃𝐷 = ∠𝐶𝐴𝐵 (∠𝑠 in the same segment) Hence, △CDX and △CBA are similar. 6 (i) DT is parallel to AB. (Midpoint Theorem) ∠𝐴𝐹𝐷 = ∠𝑇𝐷𝐹 (alt angles) = ∠𝐹𝐸𝐷 Since ∠𝐴𝐹𝐷 and ∠𝐹𝐸𝐷 satisfies the alternate segment theorem, AB is a tangent at F. (ii) ∠𝐷𝐹𝐸 is common. ∠𝑇𝐷𝐹 = ∠𝐷𝐶𝐹 (base angles of an isos triangle) ∠𝐷𝐶𝐹 = ∠𝐷𝐸𝐹 (angles in the same segment) ∴ 𝐷𝐹𝑇 and EFD are similar triangles (AA) 𝐷𝐹 𝐸𝐹 = 𝐹𝑇 𝐹𝐷 𝐷𝐹2 = 𝐹𝑇 × 𝐸𝐹 = 𝐹𝑇 × (𝐸𝑇 + 𝑇𝐹) = 𝐹𝑇2 + 𝐹𝑇 × 𝐸𝑇 𝐷𝐹2 = 𝐹𝑇2 + 𝐹𝑇 × 𝐸𝑇

Page 6 Paradigm Specialising in O Level Mathematics Maths Secrets 7 (i) ∠𝐵𝐶𝐴 = ∠𝐴𝐶𝐸 (Common angle) ∠𝐴𝐵𝐶 = ∠𝐶𝐴𝑌 (Angles in the alternate segments) = ∠𝐸𝐴𝐶 (AC bisects ∠𝐷𝐴𝑌) ∴△ 𝐴𝐵𝐶 and △ 𝐴𝐸𝐶 are similar. 𝐴𝐶 𝐸𝐶 = 𝐵𝐶 𝐴𝐶 (corresponding sides of similar triangles) 𝐴𝐶2 = 𝐸𝐶 × 𝐵𝐶 (shown) (i) ∠𝐶𝐴𝑌 = ∠𝐸𝐴𝐶 (AC bisects ∠𝐷𝐴𝑌) ∠𝐵𝐴𝑋 = ∠𝐸𝐴𝐵 (AB bisects ∠𝐵𝐴𝑋) ∠𝐵𝐴𝑋 + ∠𝐸𝐴𝐵 + ∠𝐸𝐴𝐶 + ∠𝐶𝐴𝑌 = 180˚ (angles on a straight line) 2∠𝐸𝐴𝐵 + 2∠𝐸𝐴𝐶 = 180˚ ∠𝐸𝐴𝐵 + ∠𝐵𝐴𝐶 = 90˚, BC is a diameter of the circle. (ii) ∠𝐴𝐵𝐸 = ∠𝐶𝐴𝑌 (Angles in the alternate segments) ∠𝐶𝐴𝑌 = ∠𝐸𝐴𝐶 (AC bisects ∠𝐵𝐴𝑌) ∴∠𝐴𝐵𝐸 = ∠𝐸𝐴𝐶 ∠𝐸𝐴𝐵 + ∠𝐸𝐴𝐶 = ∠𝐸𝐴𝐵 + ∠𝐴𝐵𝐸 = 90˚ (from (ii)) ∠𝐴𝐸𝐵 = 90˚ (sum of ∠𝑠 in a triangle) ∴ 𝐴𝐷 and 𝐵𝐶 are perpendicular. 8 (i) To prove: BD||FE Proof: Let ∠𝑇𝐴𝐹 be 𝜽. ∠𝐴𝐵𝐷 = ∠𝑇𝐴𝐹 = θ (alt seg thm) ∠𝐴𝐸𝐹 = ∠𝑇𝐴𝐹 = θ (alt seg thm) ∴ ∠𝐴𝐵𝐷 = ∠𝐴𝐸𝐹 = θ Using property of corresponding angles, BD||EF (shown) (ii) To prove: ∠𝐹𝐴𝐶 = ∠𝐶𝐴𝐸 Proof: Let ∠𝐵𝐶𝐸 = 𝛼 ∠𝐶𝐵𝐷 = ∠𝐵𝐶𝐸 = 𝛼 (alt ∠𝑠, BD||EF) ∠𝐹𝐴𝐶 = ∠𝐶𝐵𝐷 = 𝛼 (∠𝑠 in same segment) Also, ∠𝐶𝐴𝐸 = ∠𝐵𝐶𝐸 = 𝛼 (alt seg thm) ∴∠𝐹𝐴𝐶 = ∠𝐶𝐴𝐸 = 𝛼 (shown) 9 (i) ∠𝐶𝐴𝐵 = ∠𝐶𝐷𝐴 (Alternate Segment Theorem) And ∠𝐵𝐷𝐴 = ∠𝐶𝐷𝐴 (same angle) ∠𝐴𝐵𝐶 = ∠𝐴𝐵𝐷 (Common angle) Triangle ABD is similar to triangle CBA. (AA) (ii) ∠𝐶𝐴𝐵 = ∠𝐶𝐷𝐴 (Alternate Segment Theorem) ∠𝐶𝐴𝐵 = ∠𝐴𝐶𝐻 (Alternate angles, BAB||FEHC) Hence ∠𝐴𝐶𝐻 = ∠𝐶𝐷𝐴 ∠𝐻𝐴𝐶 = ∠𝐷𝐴𝐶 (Common angle