Question

In the given fig. ABCD is a parallelogram. A circle through A, B and C intersects CD produced at E. Prove that AD = AE.

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Answer 1

ABCE IS A CYCLIC QUADRILATERAL
and ABCD is a parallelogram
As AB║CD or AB║CE
ABCE can be called a trapezium
and a cyclic trapezium is an isosceles trapezium 

∴AE = BC___________(1)
and as AD =BC

∴AD = AE {FROM ()}

Answer 2

given : DC ║ AB
so, EC ║ AB
also, AD = BC 
and ∠B + ∠C = 180°
but, ABDE is a cyclic quadrilateral so,
∠B + ∠E = 180°
⇒ ∠E = ∠C
Now, draw AL ⊥ EC & BM ⊥ EC
In tri. ALE & BMC,
 ∠E = ∠C
 AL = BM (since the distance between ║ lines is equal)
∠ALE = ∠BMC (90°)
so by AAS criterion both are congruent.
⇒  AD = AE ( by C.P.C.T.)