Question

In the given fig.ABC is a triangle in which AB=AC.If D be a point on BC produced,Prove that AD>AC

Answer 1

Given, D is a point on side BC such that AB=AC.

$\underline {To\:proof:-}$AD > AC

$\underline{Proof:-}$

In ∆ABC,

∠ACD = ∠B + ∠BAC

∠ACB + ∠BAC [ Since, AB = AC so, ∠C = ∠B]

∠CAD + ∠CDA +∠BAC [ Since, ∠ACB =

∠CAD + ∠CDA]

∴∠ACD > ∠CDA

[we know that side opp. to greater angle is longer]

∴AD > AC

Answer 2

Step-by-step explanation:

Given, D is a point on side BC such that AB=AC.

Toproof:−

AD > AC

Proof:−

In ∆ABC,

∠ACD = ∠B + ∠BAC

∠ACB + ∠BAC [ Since, AB = AC so, ∠C = ∠B]

∠CAD + ∠CDA +∠BAC [ Since, ∠ACB =

∠CAD + ∠CDA]

∴∠ACD > ∠CDA

[we know that side opp. to greater angle is longer]

∴AD > AC