Question

Assertion :In any $△ABC,$ the square of the length of the bisector $AD$ is $bc(1-\frac{a^2}{{(b+c)}^2})$ Reason: In any triangle $ABC$ length of bisector $AD$ is $\frac{2bc}{b+c}\cos \left(\frac{A}{2}\right)$

• A. Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion
• B. Both Assertion and Reason are individually correct but Reason is not the correct explanation of Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is A Both Assertion and Reason are individually true and Reason is the correct explanation of Assertion

Area of $△ABC=$ Area of $△ABD+$Area of $△ACD$
$\Rightarrow \frac{1}{2}bc\sin A=\frac{1}{2}.c.AD\sin \left(\frac{A}{2}\right)+\frac{1}{2}.b.AD\sin \left(\frac{A}{2}\right)$
$\Rightarrow bc\times 2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)=(c+b).AD.\sin \left(\frac{A}{2}\right)$
$\Rightarrow AD=\frac{2bc\cos \left(\frac{A}{2}\right)}{b+c}$
Also, ${AD}^2={\left(\frac{2bc\cos \left(\frac{A}{2}\right)}{b+c}\right)}^2$
$=\frac{4b^2{c}^2{\cos }^2\left(\frac{A}{2}\right)}{{(b+c)}^2}$
$=\frac{4b^2{c}^2}{{(b+c)}^2}\times \frac{s(s-a)}{bc}$
$=\frac{bc.2s(2s-2a)}{{(b+c)}^2}$
$=bc.\frac{(a+b+c)(b+c-a)}{{(b+c)}^2}$
$=\frac{bc[{(b+c)}^2-a^2]}{{(b+c)}^2}$
$=bc[1-\frac{a^2}{{(b+c)}^2}]$