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.
Thank you for your help.
Hello,
Why ?Forget about it.
The meeting point of the angle bisectors in a triangle is the center of the inscribed circle. Remember it for later !Originally Posted by PvtBillPilgrim
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 :
So if we find the center of this circle, we'll get A.The hypotenuse (OX1) of a right angled triangle is a diameter of the circumscribed circle.
Now, how can we find the center ?
We'll use this property :
This midpoint will be the center of the circle since OX1 is a diameter.The line bisector is constructed as follow : Line Bisector -- from Wolfram MathWorld and bisects the segment in its midpoint
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 !
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