Question

Internal bisector of angle A of triangle $ABC$ meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and the side AB at F. If a, b, c represent sides of $△ABC$, then

• A. AE is H.M. of b and c
• B. AD$=\frac{2bc}{b+c}\cos \frac{A}{2}$
• C. EF$=\frac{4bc}{b+c}\sin \frac{A}{2}$
• D. the triangle AEF is isosceles.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A AE is H.M. of b and c
B AD$=\frac{2bc}{b+c}\cos \frac{A}{2}$
C EF$=\frac{4bc}{b+c}\sin \frac{A}{2}$
D the triangle AEF is isosceles.

Given, internal angle bisector of A meets BC at D,
we know length of AD is given by, AD = $\frac{2bc}{b+c}\cos \frac{A}{2}$
Given, $△$ADE is right angled triangle with right angle at D.
$\angle D={90}^{\circ },\angle AED=90-\frac{A}{2}$
Using sine rule in $△$ADE, we get
$\frac{AD}{\sin \angle AED}=\frac{AE}{\sin \angle ADE}=\frac{DE}{\sin \angle \frac{A}{2}}$
$\frac{AD}{\cos \frac{A}{2}}=AE=\frac{DE}{\sin \frac{A}{2}}$
Substitute the value of AD in above equation, we get
AE=$\frac{\frac{2bc}{b+c}\cos \frac{A}{2}}{\cos \frac{A}{2}}=\frac{2bc}{b+c}$
so, AE is harmonic mean of b and c.
Consider $△$ADE and $△$ADF , we know that
$\angle ADE={\angle ADF=90}^{\circ }\angle DAE=\angle DAF=\angle \frac{A}{2}\therefore \angle AED=\angle AFD=90-\angle \frac{A}{2}$
Hence $△$AEF is isosceles triangle.
$EF=2DE=2AE\sin \frac{A}{2}$
$EF=2\times \frac{2bc}{b+c}\sin \frac{A}{2}=\frac{4bc}{b+c}\sin \frac{A}{2}$