Here's a geometric proof of the Pythagorean Theorem using similar triangles :
Let ABC be a righ triangle with a right angle at C
And CD is an altitude with right angles ADC and BDC
We want to prove that AC^2 + BC^2 = AB^2
So...looking at triangles ABC and ADC
Angle A = angle A
Angle ACB = angle CDA
So...by AA congruency.....Δ ACB ~Δ ADC → AC /AB = AD/ AC → AC^2 = AB * AD
Similary Δ ACB ~ ΔCDB → AB / BC = BC / DB → BC^2 = AB * DB
So
AC^2 + BC^2 = AB* AD + AB * DB
AC^2 + BC^2 = AB ( AD + DB)
AC^2 + BC^2 = AB (AB)
AC^2 + BC^2 = AB^2