ABCD is a parallelogram.The circle through A,B and C intersect CD(produced, if necessary) at E.Prove that AE=AD. with figure
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Given ABCD is a parallelogram passing through A, B, C and intersecting CD at E.
∠B = ∠D [Opposite angles of a parallelogram are equal] → (1)
Since ABCE is a cyclic quadrilateral.
Recall that “exterior angle of a cyclic quadrilateral is equal to interior opposite angle”.
Hence ∠B = ∠AED → (2)
From (1) and (2), we get
∠D = ∠AED
Therefore AE = AD [Since in a triangle, sides opposite to equal angles are equal]
∠B = ∠D [Opposite angles of a parallelogram are equal] → (1)
Since ABCE is a cyclic quadrilateral.
Recall that “exterior angle of a cyclic quadrilateral is equal to interior opposite angle”.
Hence ∠B = ∠AED → (2)
From (1) and (2), we get
∠D = ∠AED
Therefore AE = AD [Since in a triangle, sides opposite to equal angles are equal]
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