Question

If $\alpha ,\beta ,\gamma$ are the lengths of internal bisectors of the angles of a triangle ABC, then prove that
$\frac{1}{\alpha }cos\frac{A}{2}+\frac{1}{\beta }cos\frac{B}{2}++\frac{1}{\gamma }cos\frac{C}{2}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Determine the lengths of medians also.

Answer 1

$△ABC=△ABD+△ADC$
$\frac{1}{2}bc\sin A=\frac{1}{2}c\alpha \sin \frac{A}{2}+\frac{1}{2}b\alpha \sin \frac{A}{2}$
Divide by $\alpha bc\sin \left(A/2\right)$
$\frac{1}{\alpha }\cos \frac{A}{2}=\frac{1}{2}(\frac{1}{b}+\frac{1}{c})$
or $\alpha =\frac{2bc}{b+c}\cos \frac{A}{2}$
Write similar expressions and add.
Lengths of medians:
If D be the mid-point, then we know that
$b^2+c^2=2(B{D}^2+A{D}^2)=2(\frac{a^2}{4}+A{D}^2)$
$\frac{2b^2+2c^2-a^2}{4}=A{D}^2$
or $A{D}^2=\frac{b^2+c^2+2bc\cos A}{4}$
$\therefore =\frac{1}{2}\sqrt{b^2+c^2+2bc\cos A}$etc.
similar expressions for medians BE and CF.