Question

. Prove that the circle drawn on

any equal side of an isosceles triangle

as diameter bisects the third

side​

Answer 1

$\huge\bold{\mid{\red{\mathfrak{\underline{Answer}}}}}$

$\large{\underline{\bf{Explanation}}}$

★Draw ABC as an Isosceles ∆.

★On the side AB mark the mid point O.

★Now O as the centre,draw a circle with radius =OB=OA.

★It may not intersect base of all Isosceles ∆s'.

★But we choose base BC of our circle LONG enough so that it will intersect BC at D.

Now,

OB=OA=OD=Radius

AB=2*R=AC {Isosceles ∆}

In triangle OBD,

Angle B=Angle D as sides are equal.

Since

Angle B=Angle C

So therefore,

Angle B=Angle C=Angle D

$\bigcirc$∆ OBD and ABC are similar.

$\bigcirc$AB ||OB,BD||BC

$\bigcirc$ All Angles are Equal.

_________________________________

OB=1/2 AB

BD=1/2 BC

‡Hence Proved‡