ABCD is a cyclic quadrilateral in which AB and CD when produced meet at E and EA=ED. Prove that AD is parallel to BC and EB=EC. Sir please tell me two ways for doing this question.

Asked by Nitin Joshi | 24th Nov, 2013, 09:50: AM

Expert Answer:

1) In EAD,

EA = ED

Let ∠EAD = x

Thus, ∠EAD =∠EDA = x  [angles opposite to equal sides are equal]

∠BCD +∠DAB =1800

∠BCD + x = 180 [since ∠DAB =∠DAE = x]

∠BCD = 180 - x

Similarly,

∠ABC = 180 - x

∠DAB +∠ABC = x + 180 - x

∠DAB +∠ABC =1800  

Similarly, ∠BCD + ∠CDA = 1800

Since the adjacent interior angles are supplementary, therefore

2) Since ,

∠EBC =∠EAD = x [corresponding angles are equal]

Similarly ∠ECB =∠EDA = x

Now, in triangle EBC,

∠EBC =∠ECB = x

Thus, EB = EC     [sides opposite to equal angles are equal]

Answered by Rashmi Khot | 24th Nov, 2013, 12:37: PM

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