Q. 333.5( 4 Votes )

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

Answer :

Given: ABCD is a parallelogram. The circle through A, B and C intersects CD (produced, if necessary) at E.
To Prove: AE = AD
Proof: In cyclic quadrilateral ABCE,
∠ AED + ∠ ABC = 180° ...(1)   (Since opposite angles of a cyclic quadrilateral are supplementary)

Also, ∠ ADE + ∠ ADC = 180°          (Linear Pair Axiom )
But ∠ ADC = ∠ ABC            (Opposite angles of a parallelogram )
∴ ∠ ADE +∠ ABC = 180°                  ...(2)
From (1) and (2), we have
∠ AED + ∠ ABC = ∠ ADE + ∠ ABC
⇒ ∠ AED = ∠ ADE
In triangle ADE,
AE = AD
∴ Sides opposite to equal angles of A are equal. Proved.

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