ABCD is a trapezium where AB || DC and AD = BC.
Draw DM and CN perpendicular to AB.
In Δ DAM and Δ CBN,
⇒ AD = BC (Given)
⇒ ∠AMD = ∠BNC (Right angles)
⇒ DM = CN (Perpendicular distance between parallel lines are equal)
∴ ∆ DAM ≅ ∆ CBN, (By R.H.S. congruence rule)
We know that,
Corresponding part of congruent triangle are equal.
⇒ ∠A = ∠B (By C.P.C.T.) .....(1)
From figure,
⇒ ∠B + ∠C = 180° (Sum of the co-interior angles = 180°)
Substituting value of ∠B from equation (1) in above equation, we get :
⇒ ∠A + ∠C = 180°
We know that,
Sum of angles in a quadrilateral = 360°.
∴ ∠A + ∠B + ∠C + ∠D = 360°
⇒ (∠A + ∠C) + (∠B + ∠D) = 360°
⇒ 180° + (∠B + ∠D) = 360°
⇒ (∠B + ∠D) = 360° - 180°
⇒ (∠B + ∠D) = 180°.
ABCD is a cyclic quadrilateral as the sum of the pair of opposite angle is 180°.
Hence, if the non-parallel sides of a trapezium are equal, it is cyclic.
Circles — Interactive Study Guide
Circle Theorems Quick Reference
- Equal chords ⇔ equal central angles
- Perpendicular from centre bisects the chord
- Equal chords ⇔ equidistant from centre
- Central angle = 2 × inscribed angle (same arc)
- Angles in same segment are equal
- Angle in semicircle = 90°
- Opposite angles of cyclic quad = 180°
Problem-Solving Toolkit
Quick Self-Check
- Inscribed angle = 35°. Central angle for the same arc? (70°)
- Chord = 10 cm, distance from centre = 12 cm. Radius? (√(25+144) = √169 = 13 cm)
- ABCD is cyclic, ∠A = 95°. Find ∠C. (85°)