prove that the area of a right angled triangle is equal to the product of hypotenuse parts

Question

I have a right-angled triangle $ABC$, right angle at $B$. The inscribed circle touches ${AC}$ in point $D$. Prove that $\mid AD \mid \cdot\mid DC\mid$ is equal to the area of the triangle.

Answer 1

It's true if $AC$ is a hypotenuse of the triangle.

Indeed, let $AB=c$, $AC=b$ and $BC=a$.

Thus, $$AD\cdot DC=\frac{b+c-a}{2}\cdot\frac{a+b-c}{2}=\frac{b^2-a^2-c^2+2ac}{4}=\frac{ac}{2}=S_{\Delta ABC}$$

Answer 2

Let I be the in-center, r the radius of the in-circle. The in-circle touches AB, BC at E and F respectively.

Because of tangent properties, AD = AE, BE = BF, and CD = CF.

On one hand, we have $S_{ABC} = \dfrac {AB \times BC}{2}$

$ = \dfrac {(AD + r)(DC + r)}{2} = \dfrac {AD.DC + r.AD + r.DC + r^2}{2}$

$ = \dfrac { AD.DC }{2} + \dfrac {r.AD + r.DC + r^2}{2} = \dfrac { AD.DC }{2} + k$; where $k = \dfrac {r.AD + r.DC + r^2}{2}$

That is, $S_{ABC} = \dfrac { AD.DC }{2} + k$ .... (1)

On the other hand, we have $S_{ABC} = 2 \times S_{IAD} + 2 \times S_{IDC} + S_{IEBF}$

$ = 2 \times \dfrac {r.AD}{2} + 2 \times \dfrac {r.DC}{2} + r^2 = r.AD + r. DC + r^2 = 2k$

That is, $ S_{ABC} = 2k$ .... (2)

Required result follows after eliminating the k from (1) and (2).