Question

in an isosceles triangle ABC in a figure ab is equal to AC show that BF is equal to FC​

Answer 1

$\large\underline{\sf{Given- }}$

• An isosceles triangle ABC, with AB = AC

and

• A circle is inscribed in it touches it sides AB, BC & AC at E, F & G respectively.

$\large\underline{\sf{To\:prove}}$

• BF = FC

$\large\underline{\sf{Solution-}}$

We know that,

• Length of tangents drawn from external point are equal.

Now, from figure

• A is an external point and AE & AG are tangents,

so,

$\rm :\implies\:AG = AE - - - (1)$

Also,

• B is an external point and BE & BF are tangents,

so,

$\rm :\implies\:BE = BF - - - (2)$

Also,

• C is an external point and CF & CG are tangents,

so,

$\rm :\implies\:CF = CG - - - (3)$

Now,

• According to statement, it is given that

$\rm :\longmapsto\:AB \: = \: AC$

• can be rewritten as

$\rm :\longmapsto\: \cancel{AE} + BE = \cancel{AG} + CG \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:BE = CG$

$\rm :\implies\:BF = CF \: \: \: \{using \: (2) \: and \: (3) \}$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

1. The tangent touches the circle at only one point

2. It is perpendicular to the radius of the circle at the point of tangency

3. The length of tangents from an external point to a circle are equal.