Geometry is not that bad

Level 2

In a triangle $\triangle{ABC}$ we draw a cevian through C that touches AB at G. The angle bisector of $\triangle{BGC}$ touches BC at F and the angle bisector of $\triangle{CGA}$ touches CA at E. If you know $\overline{FG}$ = 33 and $\overline{GE}$ = 56. What is $\overline{EF}$ ?

The answer is 65.

3 solutions

Michael Ng
Dec 23, 2014

Notice that $\angle EGF = 90^{\circ}$ so by Pythagoras' Theorem $EF = \boxed{65}$ .

Some people are not in your level Michael, you should explain it in order that everybody can understand!

$\angle{EGF} = 90^\circ$ because $\angle{BGF} + \angle{FGC} + \angle{CGE} + \angle{EGA} = 180^\circ$ and because $\angle{BFG} = \angle{FGC} = \alpha$ and $\angle{CGE} = \angle{EGA} = \beta$ we have $2\alpha + 2\beta = 180^\circ \implies \alpha + \beta = 90^\circ$ and from here Michael Ng solution follows.

Jordi Bosch - 6 years, 5 months ago

Ah thank you. I was writing it in a rush :)

Michael Ng - 6 years, 5 months ago

By the way, I didn't want to sound rude, thanks Michael for bothering writing a solution!!

Jordi Bosch - 6 years, 5 months ago

thanx dude :D

madhav srirangan - 6 years, 5 months ago

Agastya Chandrakant
Dec 31, 2014

<AGC +<BGC=180° So the sum of the angles formed by angle bisector is 90°. Right angle! Cheers! Pythagoras theorem...

Anna Anant
Jan 2, 2015

<EGF is 90. Then using Pythagorean theorem, EF = sqrt(33 33 + 56 56) = 65

Geometry is not that bad

The answer is 65.

3 solutions

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