Prove this: (1 Viewer)

[Dreamerish*~]

This question's driving me mental:

BX and BY are the interior and exterior bisectors of angle ABC. Prove that BX and BY are perpendicular.

 

[jessied]

Does it come with a diagram?

 

[Dreamerish*~]

It does now.

 

[mojako]

answer attached
u can sort of guess what exterior bisector means when they tell u it's perpendicular to interior bisector
also it's like the bisector of the exterior angle in a triangle

 

[Dreamerish*~]

Thanks.

I guess my definition of the exterior bisector is wrong.

 

[acmilan]

Exterior bisector basically bisects the supplementary angle of the angle in question. So you just extend one of the sides and bisect that angle formed.

 

[jessied]

I still have no idea even after seeing the answer, but meh...

Where'd the question come from?

 

[jessied]

Exterior bisector basically bisects the supplementary angle of the angle in question. So you just extend one of the sides and bisect that angle formed.

Oh I get it now..

 

[Dreamerish*~]

Exterior bisector basically bisects the supplementary angle of the angle in question. So you just extend one of the sides and bisect that angle formed.

Thank you.

Yeah, I thought it was bisecting by creating another angle of the same size outside the original one, if that makes sense.

It's from some Extension 1 revision exercises in the back of a random textbook. I still don't know what it's called.

 

[lucifel]

it's actually not too bad.

Ok assume the triangle is drawn ABC in anti clock wise direction.

Assume BX bisects the ext angle. And BY the interior.

So let /_ABC = 2x
So /_ABY = x (BY bisects/_ABC)
Now ext /_ = 180-2x (/_sum st line = 180)
so /_ABX = 90-x (BX bisects ext/_)
So /_XBY = 90-x +x
= 90
thus, BY and BX are perpindicular.

It's poorly type set, but I can't be stuffed openning Latex, so yeh.

Oh my in the time it took me to type all that, the question had already been answered, and in a way better than mine. Whoops. Oh well.