Question

In Fig., if $AB=AC$, prove that $BE=EC$ [2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

Since tangents from an exterior point to a circle are equal in length.

Therefore, $AD=AF$ ...(i) [Tangents from A]

$BD=BE$ ...(ii) [Tangents from B]

$CE=CF$ ...(iii) [Tangents from C]

Given, $AB=AC$

$\Rightarrow AB–AD=AC–AD$

[Subtracting AD from both sides]

$\Rightarrow AB–AD=AC–AF$ [Using (i)]

$\Rightarrow BD=CF\Rightarrow BE=CF$ [Using (ii), BE = BD]

$\Rightarrow BE=CE$ [Using (iii), CE = CF]

Hence proved.