Ex 10.6, 6 (Optional)
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD .
Given: A Parallelogram ABCD a circle is drawn through A, B and C
which intersects CD Produced at E.
To Prove: AE = AD
Construction: Join AE
Proof:
Since points A, B, C, E lie on a circle
∴ ABCE is a cyclic quadrilateral.
We know that,
Sum of opposite angles of a cyclic quadrilateral is 180° .
Therefore,
∠AEC + ∠ABC = 180°
Also,
∠ADE + ∠ADC = 180°
Also,
Opposite angles of a Parallelogram are equal,
∴ ∠ADC = ∠ABC.
Putting (3) in Equation (2)
∠ADE + ∠ADC = 180°
(By Linear Pair)
∠ADE + ∠ABC = 180°
Now, our equations (1) and (4) are
∠AEC + ∠ABC = 180° …(1)
∠ADE + ∠ABC = 180° …(4)
Comparing (1) and (4)
∠AEC + ∠ABC = ∠ADE + ∠ABC
∠AEC = ∠ADE
∠ AED = ∠ADE
AE = AD
Hence Proved
(Sides opposite to equal angles are equal)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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