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'?

There are only two possible pairs of right triangles that can share a common side to form a triangle with a perimeter less than 50 and all sides of integer length. These are:
[(12, 6, 13), (12, 9, 15)], base length = 15, perimeter = 43; and
[(8, 6, 10), (8, 15, 17)], base length = 21, perimeter = 48. 

I count four possible pairs of triangles that can share a common side to form a triangle with a perimeter less than 50 and all sides of integer length where the base is the longest side and where a line segment from the top vertex to the base divides the triangle so that the pair of triangles have 60° and 120° angles with the base and the line segment.  These are:
[(5, 3, 7), (5, 8, 7)], base length = 11, perimeter = 25;
[(3, 5, 7), (3, 8, 7)], base length = 13, perimeter = 27;
[(8, 7, 13), (8, 15, 13)], base length = 22, perimeter = 48; and
[(7, 8, 13), (7, 15, 13)], base length = 23, perimeter = 49.

The two triangles with a base length with a difference of 1, then would be:
triangle  ABC   (21, 10, 17), perimeter = 48; and
triangle A'B'C' (22, 13, 13), perimeter = 48.
The lengths of the line segment dividing the two triangles are:
CP = 8 and C'P' = 8.