1. ## triangle construction

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.

2. 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.

Are you familiar with the game Risk? The old pieces for the denomination of one army had three lines intersecting with smoothed intersections such as the ones in this picture.

Now, the cross section of these pieces, excluding the curvature, can represent your bisectors. Choosing a point x on any of those lines, and letting it be on the triangle makes this a simple problem. However, the triangle XYZ denotes a triangle formed my drawing lines from X to Y, Y to Z and Z to X. Therefore, X is treated as a vertex of the triangle. The bisector is the line that lies on the point exactly in the middle of two vertices. Therefore, to make this situation valid, the point X should lie in the center of the bisectors/risk-army/triangle effectively making the triangle non-existent because it has an absolute area of zero.

3. So you're saying you can't construct this? Is that right? I can use geometric transformations if that helps.

Anyway, you can definitely construct it. The intersection of the three lines is the incenter of the triangle.

I'm going to post this in the advanced geometry section.

5. Concentration... lol !

1. : introduction
This is the figure at the beginning.
The point A will be one summit of the final triangle and O is the intersection of the 3 lines.
L and M are construction points and have no importance.

2. : first symmetry
We draw the circle d of center O and radius OA. This is the circumscribed circle.
Point G is a random point on line b.

We'll construct the symmetric point E of A with respect to line b :
- since G is on this line, GA=GE. So draw the circle of center G and radius GA.
- since O is on this line, OA=OE. So E is on the circle d.
Thus E is the intersection of the two circles.

For more convenience, the circle of construction has been removed.

3. : second symmetry
Point H is a random point on line c
Point F is the symmetric point of A with respect to line c and is constructed exactly the same way as E.

Draw lines GA and GE (blue) and lines HA and HF (green)

4. : key step/explanation
Imagine EF will extend to be a side of the final triangle.

Why did we construct the symmetrics of A ? Here is the explanation :
Since A and E are symmetric wrt line b, any point G on this line will be at equal distance from A and E : AG=EG.
Therefore, triangle AGE will be isosceles (summit G).
Since line b is the axis of symmetry for A and E, it's the perpendicular bisector of line AE.
But in an isosceles triangle, the perpendicular bisector to the base is also the angle bisector of the summit angle.
Therefore, b will be the angle bisector of any angle $\angle AGE$, where G can be any point of line b.

Same goes for any point H on line c.

Do you see better now ? Ok, I understand that the figure is messy... But think hard

5. : final
Let's call B the intersection of lines EF and b and C the intersection of lines EF and c.

Since B is on line b, b bisects $\angle ABE=\angle ABC$
Since C is on line c, c bisects $\angle ACF=\angle ACB$

$\text{Therefore, in triangle ABC, lines \textbf{b} and \textbf{c} are angle bisectors}$ $\text{ and thus their intersection, O, is the } \boxed{\text{center of the inscribed circle}}$

No matter what, line OA, aka line a aka OA, will be an angle bisector.

And we're done !

6. Great.. Hats off to Moo..

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sam
FSBO