Question

14.
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.
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Answer 1

Answer:

Step-by-step explanation:

We have

Since ABC is an isosceles triangle with AB = AC and DE is parallel to BC

So

angle ADE = angle ABC. __ corresponding angles

angle SVC = angel ACB____ (opposite angles of isoseleles triangle )

=>angle ADE = angle ACB

now,

angle ADE +angle EDB = 180°

angle ACB + angle EDB =180°

Thus the opposite angles of DECB are supplementary.

Hence DECB is a cylclic quadrilateral

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Answer 2

Answer:

We have

Since ABC is an isosceles triangle with AB = AC and DE is parallel to BC

So

angle ADE = angle ABC. __ corresponding angles

angle SVC = angel ACB____ (opposite angles of isoseleles triangle )

=>angle ADE = angle ACB

now,

angle ADE +angle EDB = 180°

angle ACB + angle EDB =180°

Thus the opposite angles of DECB are supplementary.

Hence DECB is a cylclic quadrilateral

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Step-by-step explanation: