ABCD is a parallelogram. The circle through A,Band C intersect CD at E. Prove that AE=AD.
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Given: ABCD is a parallelogram. A circle through A, B, C intersects CD produced at E.
To prove: AE = AD
Proof:
∠AED + ∠ABC = 180° ... (1) (Sum of opposite angles of cyclic quadrilaterals is 180°)
∠ADE + ∠ADC = 180° ... (2) (linear pair)
∠ABC = ∠ADC ... (3) (opposite angles of parallelogram are equal)
From (1) and (2)
∠AED + ∠ABC = ∠ADE + ∠ADC
⇒∠AED = ∠ADE (using (3))
In ∆AED,
∠AED = ∠ADE
⇒AD = AE (equal sides have equal angles opposite to them)
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