Construct triangle $ABC$ given point $C$ and the lines that contain the angle bisectors of angles $A$ and $B$.

Question

The problem is as stated in the title: Construct triangle $ABC$ given point $C$ and the lines that contain the angle bisectors of angles $A$ and $B$.

I'm slightly confused about the whole constructions thing; I know we're supposed to consider the completed diagram, find some relations, and use that information to conversely construct the diagram. Are we then supposed to prove that the constructed diagram actually satisfies the given conditions? If someone could clarify that, and give a solution to the above problem illustrating the construction and proof of construction, that would be awesome.

Answer 1

Here is a way to construct $\triangle ABC$.

Find the interception point O (incenter) of angle bisectors of $\angle A$ and $\angle B$. The constructed $\triangle ABC$ should have circle O as its incircle.

Draw a circle with radius r centered at incenter O. Draw two tangent lines CD and CE to the circle that intercept two angle bisector lines at D and E. Connect DE, adjust the radius r so that DE is tangent to the circle, then $\triangle DEC$ is $\triangle ABC$.

Answer 2

Let $I=\ell_1 \cap\ell_2$, $D\in\ell_1$, $E\in\ell_2$ as in the image.

Construct $IF\perp IE$. Then $\angle DIF=\tfrac\gamma2$.