# Proving two bisectors are equal

• Jun 16th 2009, 06:44 AM
BG5965
Proving two bisectors are equal
In a triangle ABC, angle B is 60 degrees. Angles A and C are bisected to points D and E respectively, on the edge of the triangle (see diagram). They cross at point O. Prove that OD = OE

http://i301.photobucket.com/albums/n.../Noether13.jpg

I've gone through it a little and got this.

2x + 2y + 60 = 180
2x + 2y = 120
x + y = 60

I don't know if this helps at all.

Remember, angles A and C are bisected, so the two adjacent angles are equal to each other (x, x and y, y)

Thanks for any help, BG
• Jun 16th 2009, 11:19 AM
Geometry Proof
Hello BG5965
Quote:

Originally Posted by BG5965
In a triangle ABC, angle B is 60 degrees. Angles A and C are bisected to points D and E respectively, on the edge of the triangle (see diagram). They cross at point O. Prove that OD = OE

http://i301.photobucket.com/albums/n.../Noether13.jpg

I've gone through it a little and got this.

2x + 2y + 60 = 180
2x + 2y = 120
x + y = 60

I don't know if this helps at all.

Remember, angles A and C are bisected, so the two adjacent angles are equal to each other (x, x and y, y)

Thanks for any help, BG

Join $OB$, and note that it is the bisector of $\angle B$, since the angle bisectors of a triangle are concurrent.

$\Rightarrow \angle OBD = \angle OBE = 30^o$

Also $\angle EOA = x + y$ (exterior angle of $\triangle AOC$)

$\Rightarrow EOA = 60^o = \angle EBD$

$\Rightarrow EODB$ is a cyclic quadrilateral (exterior angle = interior opposite angle)

Now join $ED$, and use angles in the same segment:

$\angle OBD = \angle OED = 30^o$ and $\angle OBE = \angle ODE = 30^o$

$\Rightarrow \angle ODE = \angle OED$

$\Rightarrow EO = OD$ (isosceles $\triangle OED$)