8. EUCLIDEAN GEOMETRY 1 Recap Geometry: Basic Reasoning Recap Geometry: Parallel lines 5. THEOREM: Opposite ’s of cyclic quad 2 The opposite angles of a cyclic quadrilateral are supplementary. Given: Circle with centre O passing through the vertices of cyclic quadrilateral ABCD. Prove: A + C = 180◦ and B + D = 180◦ Proof: Draw BO and DO. 5. THEOREM: Opposite ’s of cyclic quad 3 The opposite angles of a cyclic quadrilateral are supplementary. O1 = 2 A ( at centre = 2 × at circle) and O2 = 2 C ( at centre = 2 × at circle) ∴ O1 + O2 = 2 ( A + C) 5. THEOREM: Opposite ’s of cyclic quad 4 The opposite angles of a cyclic quadrilateral are supplementary. But O1 + O2 = 360◦ (’s around a point) ∴ A + C = 180◦ A+ B + C + D = 360◦ (sum angles of a quadrilateral) Moving opposite ∴ B + D = 180◦ angles in a cyclic quad Example: Solve for x and y 5 x + 76 ◦ = 180 ◦ (opp. ’s of a cylic quad) x = 104 ◦ y + 82 ◦ = 180 ◦ (opp. ’s of a cylic quad) y = 98 ◦ 6. THEOREM: Tangents from same point 6 Two tangents drawn to a circle from the same point outside the circle are equal in length. Given: Circle with centre O and tangents PA and PB touching the circle at A and B. Prove: PA = PB Proof: Draw OA, OB and OP. 6. THEOREM: Tangents from same point 7 Two tangents drawn to a circle from the same point outside the circle are equal in length. In ∆ PAO and ∆ PBO: i) PAO = PBO (tangent radius) ii) PO is a common side iii) AO = BO (radii equal) ∴ ∆ PAO ≡ ∆ PBO (RHS) ∴ PA = PB Example: Solve for d, given HG and HI are tangents to the circle 8 HG = HI (tangents from same point) HG = 8 cm JGH = 90 ◦ (radius tangent) d2 + JG2 = HG2 d2 = (8)2 – (5)2 (Pythagoras) d = 6,24 cm 7. THEOREM: Tan-chord theorem 9 The angles between a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segment. Given: Circle with centre O; SBR a tangent at B; AB a chord and points Q and P on the major and minor arcs. Prove: i) ABS = Q ii) RBA = P Proof: Draw BT & AT. 7. THEOREM: Tan-chord theorem 10 The angles between a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segment. TBA + ABS = 90◦ (tangent diameter) TAB = 90◦ ( in semi-circle) ∴ TBA + T = 90◦ (’s of ∆ TAB) 7. THEOREM: Tan-chord theorem 11 The angles between a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segment. ∴ ABS = T But, T = Q (’s in same segment) ∴ ABS = Q Hence, (i) ABS = Q 7. THEOREM: Tan-chord theorem 12 The angles between a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segment. RBA + ABS = 180◦ (’s on a straight line) and Q + P = 180◦ (Opposite ’s of a cyclic quad) ∴ RBA + ABS = Q + P 7. THEOREM: Tan-chord theorem 13 The angles between a tangent to a circle and a chord drawn from the point of contact are equal to the angles in the alternate segment. But ABS = Q (proved in (i)) Hence, (ii) RBA = P Example: Solve for i, j and k 14 i + 39 ◦ + 101 ◦ = 180 ◦ (’s on a straight line) i = 40 ◦ k = i (tan-chord theorem) k = 40 ◦ j = 101 ◦ (tan-chord theorem) Tan-chord Theorem Examples Summary Example: Solve for a, b & c 15 a + 64 ◦ = 90 ◦ (radius to tangent) a = 26 ◦ b = 64 ◦ (tan-chord theorem) c = 2 × b ( at centre = 2 × at circle) c = 128 ◦ Summary & Examples of Circle Theorems .