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.

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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