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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I hv seen this type of ques in which d was outside the figure and e was on the circumference
you have written ABED as a cyclic quadrilateral
jzt correct it as ABCE
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Answers

2016-03-02T11:31:26+05:30
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 ()}

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The Brainliest Answer!
2016-03-02T11:31:39+05:30
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.)

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It is not given that ABCE is a cyclic quadrilateral.
but we can see tht in the figure
but their points are concylic therefore ABCE is cyclic
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