Question

D is a point on the side BC of a $△ABC$ such that AD bisects $\angle BAC$. Then

• A. $BD=CD$
• B. $BA>BD$
• C. $BD>BA$
• D. $CD>CA$

Answer 1

Answer: (B)

$BA>BD$

Given-

AD is the angular bisector of a triangle ABC.

AD meets BC at D.

To find out-

which of the given options is true.

Solution-

AD is the bisector of $\angle BAC.$

$\therefore \angle BAD=\angle CAD.$

Again , in $\Delta ACD,\angle ADC$is the external angle .

We know that the external angle of a triangle is greater than each intrenal opposite angle of the same triangle.

$\therefore \angle ADB=\angle DAC+DCA.$

$=\angle BAD+\angle DCA(\because \angle DAC=\angle BAD)i.e\angle ADC>\angle BAD.$

$\therefore BA>BD.$

So option B is true.

Option A is true only when the triangle is equlateral or isosceles.

Option C is false since Option B is true.

option D is false since

$\angle ADC>\angle BADor\angle CAD$

$\therefore CD\ngtr CA$\

Ans- Option B.