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  • If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.    

Answers (1)

Given: A line is drawn parallel to the base of an isosceles triangle to intersect its equal sides

Let \trianglePQR be an isosceles triangle such that PR = PQ

Also, RQ || MN

To prove: Quadrilateral MNQR is a cyclic quadrilateral.

Proof:

In DPQR

PR = PQ [Given]

\Rightarrow \anglePQR = \anglePRQ … (1)  (angles opposite to equal sides in a triangle are equal)

Now, RQ || MN and PQ is a transversal

\Rightarrow \anglePNM = \anglePQR … (2)  (Corresponding angles)

On adding \angleMNQ to both sides in equation (2), we get:

\anglePNM + \angleMNQ = \anglePQR + \angleMNQ

180° = \anglePQR + \angleMNQ         (\anglePNM + \angleMNQ forms a linear pair)

\Rightarrow \angleMNQ + \anglePRQ = 180°    (From equation (1))

Hence, MNRQ is a cyclic quadrilateral as the sum of opposite angles is 180o

Hence proved

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