Using Pythagoras' theorem,
$\displaystyle \left(a_1\right)^2+h^2=a^2$
$\displaystyle a^2+b^2=\left(a_1+b_1\right)^2$
$\displaystyle \left(b_1\right)^2+h^2=b^2$
Therefore
$\displaystyle \left(a_1\right)^2+h^2+\left(b_1\right)^2+h^2=\lef t(a_1+b_1\right)^2$
from which the result follows
Yes,
as the non-right-angles are acute, we have
$\displaystyle a_1b_1=h^2\Rightarrow\frac{a_1b_1}{h}=h\Rightarrow \frac{b_1}{h}=\frac{h}{a_1} $
These ratios are the tangents of two angles, which being equal causes the angles to be equal.
Both inner triangles have a right-angle, therefore the third angles are also equal
and the two acute angles sum to 90 degrees.
Hi guys,
Solution 0f followup question using Pythagorean Theorem and same side ids given h^2 =a1 x b1
h^2 + a1^2 =a^2
h^2 + b1^2 =b^2
2h^2 + a1^2 + b1^2 =a^2 + b^2
2a1xb1 + a1^2 + b1^2 = a^2 +b^2
(a1 + b1 )^2 = a^2 + b^2
triangle is a right triangle