Prove BC.DE = AC.AE + AB.AD

Question

Let ABC be a triangle. On the opposite ray of the ray AC take point D, on the half-plane of the DC edge that does not contain B draw ray Dx such that CDx  equals ABC. Let E be the intersection of rays Dx and AB.

Prove BC.DE = AC.AE + AB.AD

Comments

Your description doesn't make sense. Please provide a picture.

"Opposite ray" has no meaning even if "ray AC" means the side AC (extended) of the triangle ABC.

Half plane suggests a three-dimensional structure, whereas other parts of the question suggest a plane figure. Which is it? 3-D or 2-D?

x appears to be a point which is a vertex of another triangle CDx. Does "CDx equals ABC" imply two congruent triangles, or triangles which have equal area? Or is it the measure of two angles in these triangles? If congruent, what are the corresponding sides, since there's only one common vertex (C)?

If triangles ABC and CDx are in different planes, are these planes orthogonal? Are these triangles in the same plane?

Answer 1

I think this picture may represent the figure described in the question.

The triangle ABC is shown in red. The point D is outside the triangle and X and D have been positioned so as to make the triangles ABC and CDX congruent (assumed from the question). The two angles at D and B (shaded grey) are equal, and the side pairs below are congruent:

CD=AB, DX=BC, CX=AC.

The green segments are parts of the rays DX and AB.

TO PROVE BC.DE=AC.AE+AB.AD

This is the same as proving DX.DE=CX.AE+CD.AD.

It would appear that the figure is not the intended figure implied in the question, because the proposed equation is not satisfied. A more accurate description or a picture is needed.