Question

please solve it...,..Q.no.49

Answer 1

Let DAE be tangent at A to the circumcircle of ∆ABC
In ∆ABC, AB=AC (given)
therefore, angle ABC=angle ACB -(1) [angles opposite to equal sides are equal]
According to alternate segment theorem
DAE is the tangent and AB is the chord
therefore, angle DAB=angle ACB -(2)
from (1) and (2) we have angle ABC = angle DAB
DE is parallel to BC [if a transversal intersect two lines such that a pair of alternate interior angles are equal the two lines are parallel]
therefore, the tangent at A to the circumcircle of ∆ABC is parallel to BC.

Answer 2

Simple proof for Q49.

For an Isosceles triangle ΔABC (with AB=AC), we know from its properties that the orthocenter H, circumcenter O, incenter I, center of gravity G lie on the altitude AD itself.  Altitude AD is also the median from A onto BC. 

The above points can be inferred from the symmetry of ΔABC about AD, as AB= BC, ∠B = ∠C.

Now    O lies on AD.   OA is the radius.
Tangent at A is perpendicular to radius OA (and hence to AD).
Altitude AD is perpendicular to  BC.

So the tangent is parallel to BC.