Question

Abc is an isosceles triangle with side ab =ac d is the mid point of side ac ,a circle is drawn taking bd as diameter . The point of intersection of circle and the side ab of the triangle is e .prove that ae=1/4ac

Answer 1

There's a triangle ABC with AB=AC , given
Taking BD as diameter and its mid point as O, a circle of radius BO is drawn.
E is the point where the circle touches AB
BE and DE are joined.
Angle BED must be 90 degree because it is subtended by the diameter.
We know radius bisects angle.
In triangle ODE,
Angle ODE = 45 degree. We know, two radii form an isosceles triangle.

Hence triangle ODE is an isosceles triangle with OE=OD
By angle sum property we can find out Angle DOE which comes 90 degree.
EOD + EAD = 180 (Opposite angles of quadrilateral are supplementary)
EAD = 90 degree.
AC is a tangent. Angle ODA is 90 degree (Radius is perpendicular to tangent)
But EDO = 45, ADE = 90-45 = 45 degree
In triangles AED and ODE
Angle EAD = EOD (proved earlier)
DE =DE (Common)
ADE = DEO (Proved ealier) (Therefore EO is parallel to AC)
Triangle AED is corresponding to EOD

AE = OE (By C.P.C.T.C)
EO is the half of diameter through O and the diameter would be parallel to AC too
Hence,
EO = 1/2 of diameter and diameter = 1/2 of AC
EO= 1/4 of AC
Thus AE = 1/4 of AC (Since OE =AE)