Question

prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Answer 1

Let ABC is the isosceles triangle where AB = AC
The circle drawn on AC as diameter intersects BC at D
From Circle theorem we know that the angle inscribed on semicircle is 90 degree
Hence angle ADC is 90 degree
So we have angle ADB is 90 degree
This gives Δ ADC and Δ ADB are right angled triangle.
We have hypotenuse =AB=AC and AD common leg
Two right triangles are congruent if the hypotenuse and one corresponding leg are equal in both triangles.
Hence Δ ADC ΔADB are congruent
This gives BD=DC

Answer 2

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.

  Hence the proof is done.