Cool Geometry!

Geometry Level 5

A obtuse $\Delta ABC$ is made with one of the angles equal to $120^\circ$ .

$\Delta DEF$ is made by joining the feet of the angle bisectors of $\Delta ABC$ .

Two sides of $\Delta DEF$ (in no particular order) are each $100$ units in length.

The third side is $x$ units. Find magnitude of $x^{2}$ ?

The answer is 20000.

1 solution

Ajit Athle
Apr 5, 2014

The triangle formed by the feet of angle bisectors is a right angled triangle when one of the angles of the triangle is 120°. Hence x^2 =100²+100²=20000 @Neel Khare -- It's easy to prove that /-FDE is a right angle. Extend BA to X. We see that /-DAC=/-CAX=60° or that AC is the external bisector of /-BAD. BE which bisects /-B meets AC in E. Hence E is the ex-centre of Tr.ABD. In short, DE bisects /-ADC. ///ly DF bisects /-ADB which, in turn, means that /- FDE is a right angle.

Yeah, but you missed the main part of the solution... to prove that it's a right triangle..

Satvik Golechha - 6 years, 10 months ago

but how to prove it is a right angled triangle?

Kriti Goel - 6 years, 10 months ago

@Kriti Goel , Use the exterior angle bisector theorem.

Satvik Golechha - 6 years, 8 months ago

@Kriti Goel . http://www.mathmaa.com/answers.html

A Former Brilliant Member - 4 years, 4 months ago

Cool Geometry!

The answer is 20000.

1 solution

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