Question

D is a point on side BC of $\Delta$ABC such that AD$=$AC then AB $>$ AD.

• A. True
• B. False

Answer 1

The correct option is A True

$AD=AC\left(given\right)$

$\therefore \left\{ADC\right\}!=\left\{ACD\right\}!$ [isosceles $\Delta$ ]

$\Rightarrow \left\{ADC\right\}!$ is an acute angle

[two angles of a triangle cannot be obtuse]

$\left\{ADB\right\}!={180}^0-ADC$

Which : obtuse

In $△ADB,$

$\left\{ADB\right\}!>B!$

$\Rightarrow AB>AD$

[side opposite to greater angle is greater]