Question

Prove in any $\Delta ABC$, if one angle is ${120}^o$, the $\Delta$ formed by the feet of the angle bisectors is right angled.

Answer 1

Let Triangle ABC, $\angle A={120}^{\circ }$

Let AD, BE, CF are angle bisectors of A, B, C respectively.

In $\Delta ABD$, AC is exterior angles bisector (as $\angle DAC=\angle CAX={60}^{\circ })$.

BE is internal angle bisector meet at E . E is ex-center of $\Delta ABD$

So DE is exterior angle bisector of $\Delta ABD$

Then, $\angle ADE=\angle EDC,\angle ADC=2\angle ADE.$

Similarly we can prove that for the $\Delta ABC$, F is ex-center as AB is exterior angle bisector and CF is interior angle bisector.

Hence DF is exterior angle bisector.

$\angle ADF=\angle FDB,\angle BDA=2\angle ADF$

BC is straight angle.

$\angle BDA+\angle ADC={180}^{\circ }$

$2\angle ADF+2\angle ADE={180}^{\circ }$

$\angle ADF+\angle ADE={90}^{\circ }$

Hence DEF is right triangle.

Therefore the triangle formed by the feet’s of angle bisectors of triangle whose one angle measure is ${120}^{\circ }$ is Right Triangle.