prove that circle drawn on any equal side on an isosceles triangle as diameter bisects the base
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let the circle you draw side on AB of triangle ABC intersects the base BC at a point E
and the center of the circle you draw be O . then
OB=OE=AB/2
so triangle OBE is isosceles and
angle OBE = angle ABC
there fore triangles ABC and OBE are similar.
but (OB/AB)=1/2
therefore (BE/BC)=1/2
there fore E divides BC into two
hence proved....
The diagram is given in the pic below
and the center of the circle you draw be O . then
OB=OE=AB/2
so triangle OBE is isosceles and
angle OBE = angle ABC
there fore triangles ABC and OBE are similar.
but (OB/AB)=1/2
therefore (BE/BC)=1/2
there fore E divides BC into two
hence proved....
The diagram is given in the pic below
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