Triangle ABC has an altitude drawn from C to AB, meeting the base at point P. This altitude divides the triangle into two unequal right triangles.

Triangle A'B'C' also has a point, P' on its base, with a line segment connecting it to vertex C', but chosen so that angle A'P'C' is 60°, with the resulting triangle A'P'C' though, not being equilateral.

All eight line segments are of integer length, and each triangle has a perimeter less than 50. The bases, AB and A'B', are the longest sides in each of the two respective original triangles, and they differ by 1 unit in length.

• What are the dimensions of the triangles ABC and A'B'C'?
• What are the lengths of CP and C'P'?

AB=AP+BP=√(CA²-CP²)+√(CB²-CP²)
Given CA,CP,AB and BC are integers, AB and BP
should be integers.
So, we should search for Pythagorean triplets with
a common value(CP).
The only triplets that satisfy all the conditions are
(8,15,17) and (6,8,10)
AB=21,BC=17,AC=10 or AB=21,BC=10,AC=17