How to express the area of a right triangle in terms of altitude and bisector of its right angle?

Question

In a right triangle $ABC$ with the right angle $C$, the altitude $CH$ is equal to $h$, and the angle bisector $CL$ is equal to $l$. Find the area of triangle $ABC$.

My attempt: Let $AC=a$, $BC=b$, and $AB=c$. The altitude $CH=h$ from the right angle can be found by the formula $h=\frac{ab}{c}$. The area $S$ of the right triangle can be found by $S=\frac12ab \Rightarrow ab=2S$. Then, $ch=2S$.

The bisector $CL=l$ can be found by the formula $l=\sqrt{2}\cdot\frac{ab}{a+b}\Rightarrow l(a+b)=ab\sqrt{2}\Rightarrow 2S\sqrt2=l(a+b)\Rightarrow S=\frac{l(a+b)}{2\sqrt2}$.

No idea how to proceed from here, or whether there is a more laconic solution through geometric constructions.

Answer 1

Let $\alpha$ denote the angle between the line segments of length $~l~$ and $~h~$.

Since $~l~$ and $~h~$ are known, and since $~\displaystyle \cos(\alpha) = \frac{h}{l},~$ $~\alpha~$ is known.

Designate the legs of the original right triangle to be of lengths $~b~$ and $~a,~$ where $~b~$ represents the hypotenuse of the smaller right triangle, one of whose sides is $~h~$, and one of whose angles is $~(45^\circ - \alpha).$

Since $~h~$ is known, and $~\alpha~$ is (now) known, the length $~b~$ is known.

Similarly, $~a~$ is the hypotenuse of the smaller right triangle, one of whose sides is $~h,~$ and one of whose angles is $~(45^\circ + \alpha).$

Therefore, the length $~a~$ is also known.

So, since the lengths $~a~$ and $~b~$ are known, the area of the original right triangle is known.

Answer 2

As you found,

$ h = \dfrac{ a b }{c } $

and

$ \ell = \dfrac{ a b \sqrt{2}}{ a + b } $

It follows that

$ \dfrac{1}{h^2} = \dfrac{c^2}{a^2 b^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} $

and

$ \dfrac{\sqrt{2}}{\ell} = \dfrac{1}{a} + \dfrac{1}{b} $

Square the last equation,

$ \dfrac{2}{\ell^2} = \dfrac{1}{a^2} + \dfrac{1}{b^2} + \dfrac{2}{a b} $

Subtract the previous equation,

$ \dfrac{2}{\ell^2} - \dfrac{1}{h^2} = \dfrac{2}{ab} $

Therefore, the area is given by,

$ \text{Area} = \dfrac{ab}{2} = \dfrac{1}{\dfrac{2}{\ell^2} - \dfrac{1}{h^2}} = \dfrac{ h^2 \ell^2 }{ 2 h^2 - \ell^2 } $

Answer 3

From $L$ draw $LM$ and $LN$ perpendicular to $AC$ and $BC$, respectively. The quadrilateral $LMCN$ is a square (why?) with side length $\frac{\sqrt{2}l}{2}$ (why?).

Triangles $\triangle AML$ and $\triangle LNB$ are similar to $\triangle ABC$. Therefore: $$\frac{S_{\triangle AML}}{S_{\triangle ABC}} = \frac{l^2}{2a^2} \qquad \frac{S_{\triangle LNB}}{S_{\triangle ABC}} = \frac{l^2}{2b^2}$$ (why?). Now, we have: $$S_{\triangle ABC} = S_{\triangle AML} + S_{\triangle LNB} + S_{\square LMCN} $$ Or: $$\frac{l^2}{2a^2} + \frac{l^2}{2b^2} + \frac{l^2}{2S_{\triangle ABC}} = 1$$ Or: $$\frac{1}{S_{\triangle ABC}} = \frac{2}{l^2} - \frac{1}{h^2}$$