ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE, = AD.

Ncert solutions class 9 chapter 10-37

Solution

Given

ABCD is a ||gm. The  circle through A,B and C intersect at E.

To prove:

AE=AD

Proof: 

ABCE is a cyclic quadrilateral

∠ABC+∠AEC=180°      —–[sum of opposite is of a cyclic quadrilateral is 180°]

∠AEC=180°-∠EAB  ————(i)

Now ∠AEC+∠DAE=180°-∠DAE ————(ii)

From 1 and 2

180°-∠ABC=∠180°-∠DAE

∠ABC=∠DAE   ————(iii)

Also ∠ABC=∠DAE       ————(iv)

From 3 and 4

∠ADE  =∠6

Now, In ∆ADE,

∠ADE  =∠6

∴AE=AD        ——[sides apposite to equal angles in  ∆ are equal]

Hence Proved

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