Question

Prove that circle drawn on any one of the equak sides of an isoceles triangle as diameter,bisects the third side

Answer 1

PQR is an isosceles triangle such that PQ = PR and Pl is the bisector of ∠ P.

To prove : ∠PLQ = ∠PLR = 90°

and QL = LX

In ΔPLQ and ΔPLR

PQ = PR (given)

PL = PL (common)

∠QPL = ∠RPL ( PL is the bisector of ∠P)

ΔPLQ = ΔPLR ( SAS congruence criterion)

QL = LR (by cpct)

and ∠PLQ + ∠PLR = 180° ( linear pair)

2∠PLQ = 180°

∠PLQ = 180° / 2 = 90° ∴ ∠PLQ = ∠PLR = 90°

Thus, ∠PLQ = ∠PLR = 90° and QL = LR.

Hence, the bisector of the verticle angle an isosceles triangle bisects the base at right angle.

Answer 2

Answer:

Draw ABC as an ISOSCELes triangle.  On the side  AB (lateral side) mark the mid point  O.  Now as O as the center, draw a circle with radius = OB =OA.  It may not intersect base of all isosceles triangles.  But we choose base BC of our  circle LONG enough so that it will intersect  BC (base) at D.

Now, OB = OA = OD  = radius.

AB = 2 * radius = AC (isosceles triangle)

In triangle  OBD,  anle B = angle D as sides are equal.  Since angle B = angle C, then  angle B = angle C = angle D.

triangles OBD and ABC are similar.  AB || OB,  BD || BC. and  angles are all equal.

as OB = 1/2 AB ,  BD = 1/2 BC.