ABCD is a quadrilateral in which diagonals intersect at 90 degree.
Given, AB = 6cm, CD = 5cm and DA = 6cm.
Find side BC
two isosceles triangles ... same base
Did you make your sketch?
Looks pretty straightforward to me.
Let the intersection of the diagonals be E
Look at triangle ABC, it is isosceles, so the base angles are equal.
CE is common, and we have 90° angles, so the two
triangles are congruent, making BE = ED
Now look at triangle BCD and the perpendicular CE.
Again you have a common side CE and BE = ED, so
triangles BCE and DCE are congruent ,
making BC = CD = 5