ifa line drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.PLEASE ANSWER AT THE EARLIEST!!!!
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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.
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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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