Q16 of 62 Page 10

 ABCD is a parallelogram. The circle through A, B, and C intersect CD at E (when produced). Prove that AE = AD.
 

Given: ABCD is a parallelogram and circle C(O, r) intersects CD at E.
To Prove: AE = AD
Construction: Join A to E
Proof:
In parallelogram ABCD, ∠B = ∠D (opposite angles of parallelogram) and
∠AED = ∠B          (Exterior angle of a cyclic quadrilateral is equal to the opposite interior angle)
∴ ∠ AED = ∠ ADE
∴ AE = AD (sides opposite to equal angles in a triangle).

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  15 In the figure, there are two circles intersecting each other at A and B. Prove that the line joining their centres bisects the common chord.
  17  An equilateral triangle ABC is inscribed in a circle. P is any point on the minor arc BC. Prove that PA = PB + PC.
  18  ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at right angle at M. Prove that any line passing through M and bisecting any side of the quadrilateral is perpendicular to the opposite side.