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

2

2014-12-27T19:25:03+05:30
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
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.
This gives BD=DC
2014-12-29T01:51:54+05:30

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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.

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