Question

If a line is drawn parallel to the base of an isosceles triangle which intersects the 2 equal sides of the triangle then proof that quadrilateral so formed is cyclic.

Answer 1

Step-by-step explanation:

∆ABC is isosceles triangle.  

Therefore ∠B = ∠C

A line PQ is rawn parallel to BC cutting sides AB and AC at P & Q respectively.  

We need to prove that, quadrilateral PBCQ is cyclic.  

A cyclic quadrilateral will have two sets of supplementary angles (180 degrees)  opposite.

We know that ∠B = ∠C.

PQ is parallel to BC.  

Hence ∠B + ∠P = 180 degrees. (Same side inner angles of parallel lines)

Also ∠C + ∠Q = 180 degrees.  

This also gives ∠P = ∠Q.  

We can derive from above

∠C + ∠P = 180 degrees

∠B + ∠Q = 180 degrees

Opposite angles are 180 degrees.

Hence the quadrilateral PBCQ is cyclic.