1. ## triangle

Can someone show me how to do this,

I'm given three lines L1, L2, and L3 meeting at a point and I'm also given a point X on one of the lines. Construct (with ruler and compass) the triangle XYZ with the lines L1, L2, and L3 as angular bisectors.

3. Hello,

Why ?

Originally Posted by PvtBillPilgrim
Can someone show me how to do this,

I'm given three lines L1, L2, and L3 meeting at a point and I'm also given a point X on one of the lines. Construct (with ruler and compass) the triangle XYZ with the lines L1, L2, and L3 as angular bisectors.

The meeting point of the angle bisectors in a triangle is the center of the inscribed circle. Remember it for later !

So in a first time, we have :

Now, imagine the final sketch, taking care that O is the center of the inscribed circle. (the triangle is X1 X2 X3, instead of XYZ)

Point A is the intersection of the side X1X2 with the inscribed circle. Since X1X2 is a tangent, triangle OAX1 is a right angled triangle.
Same for B.

If we can construct A, it's all done. How can we construct A ?
We'll use this property :
The hypotenuse (OX1) of a right angled triangle is a diameter of the circumscribed circle.
So if we find the center of this circle, we'll get A.

Now, how can we find the center ?
We'll use this property :
The line bisector is constructed as follow : Line Bisector -- from Wolfram MathWorld and bisects the segment in its midpoint
This midpoint will be the center of the circle since OX1 is a diameter.

The construction of the line bisector may not be clear, so to resume :
- draw circle, center X1, radius OX1
- draw circle, center O, radius OX1
- the 2 circles bisect in 2 points, which will define the line bisector.
- draw this line bisector
- call H the midpoint.

Draw circle, center H, radius OH. A & B will lie on this circle.

And this is how far I've gotten... If the triangle XYZ has to be isosceles in Y or Z, I can do something, but here, I'm stuck !