Parallelograms

Parallelograms may be replaced by rectangles with the same area and same relationship between vertices and parallel lines, so that the relationship needs only to be proven for rectangles.

Area of rectangle $ABCD=a(b+c)$

Area of rectangle $AEFG=\sqrt{a^2+b^2}\times x$

From similarity of triangles $AEB$ and $DAH$ we get $\frac yx=\frac ba$ or $y=\frac ba x$

From Pythagorean theorem in triangle $AHD$ : $x^2+y^2=x^2(1+\frac{b^2}{a^2})=(b+c)^2$

Solving for $x$ we get $x=\dfrac{b+c}{\sqrt{\frac {b^2}{a^2}+1}}$

Substituting that into the formula for area of rectangle $AEFG$ we get

$[AEFG]=\sqrt{a^2+b^2}\times\dfrac{b+c}{\sqrt{\frac{b^2}{a^2}+1}}=\sqrt{a^2+b^2}\times\dfrac{b+c}{\frac 1a\times\sqrt{a^2+b^2}}=a(b+c)$

So the areas are the same size.