Find perimeter of right $\triangle ABC$, given segments defined on sides by bisectors

Question

Given: $\triangle ABC$ is right at $A$. $BD$ and $CE$ bisect angles $\angle ABC$ and $\angle ACB$, respectively. $AE=8$ and $AD=9$

Find: perimeter of $\triangle ABC$

My solution: with the notation $x=BE$, $y=CD$, and $c=BC$, using the bisector theorem twice leads to:

$$\frac{9+y}{c}=\frac{8}{x} \ \ (1)$$ $$\frac{8+x}{c}=\frac{9}{y} \ \ (2)$$

Using pitagoras,

$$c^2=(9+y)^2+(8+x)^2 \ \ (3)$$

Then, solving the system with equations (1), (2), (3) in the unknowns $x$, $y$ and $c$. The asked perimeter will be $P=x+y+17+c$.

Question: is there another easier way to approach the problem, avoiding solving this system involving non-linear terms? Other solutions are welcomed! and sorry if this is a dup.

Answer 1

The incenter $I$ has to lie on the angle bisector of $\widehat{EAF}$ and on the locus of points $P$ such that $\widehat{EPF}=135^\circ$, which is the circumcircle of $EGF$ in the following diagram ($EG\parallel AF$):

In particular it is straightforward to locate $I$ with straightedge and compass. Then $B$ and $C$ are given by $FI\cap AE$ and $EI\cap FA$.

Answer 2

Let $|AB|=c$, $|AC|=b$, $|BC|=a$, $|AE|=u=8$ $|AD|=v=9$.

\begin{align} b&=u\cot\gamma ,\\ c&=v\cot\beta . \end{align}

\begin{align} |OA|&=|OB|=|OC|=R \\ &=\frac{b}{2\sin2\beta} =\frac{c}{2\sin2\gamma} . \end{align}

\begin{align} \frac{u\cot\gamma}{2\sin2\beta} &= \frac{v\cot\beta}{2\sin2\gamma} ,\\ \frac{u\cos\gamma}{\sin\gamma\sin\beta\cos\beta} &= \frac{v\cos\beta}{\sin\beta\sin\gamma\cos\gamma} ,\\ \frac{\cos^2\beta}{\cos^2\gamma} &=\frac{u}v ,\\ \frac{\cos2\beta+1}{\cos2\gamma+1} &=\frac{u}v ,\\ \frac{\cos2\beta+1}{\sin2\beta+1} &=\frac{u}v . \end{align}

Using identities

\begin{align} \cos2\beta &= \frac{1-\tan^2\beta}{1+\tan^2\beta} ,\\ \sin2\beta &= \frac{2\tan\beta}{1+\tan^2\beta} , \end{align}

after simplification we arrive at

\begin{align} \frac{2}{(1+\tan\beta)^2} &=\frac{u}v ,\\ \tan\beta&= \sqrt{\frac{2v}u} -1 = \frac{\sqrt{2uv}-u}u ,\\ \cot\gamma&= \cot(45^\circ-\beta) =\frac{1+\tan\beta}{1-\tan\beta} \\ \cot\gamma &=\frac{\sqrt{2uv}}{2u-\sqrt{2uv}} . \end{align}

Now we can find the sides

\begin{align} b&=u\cot\gamma =\frac{u\sqrt{2uv}}{2u-\sqrt{2uv}} ,\\ c&=v\cot\beta= \frac{uv}{\sqrt{2uv}-u} . \end{align}

For $u=8,v=9$ we have \begin{align} b&=24 ,\\ c&=18 ,\\ a&=\sqrt{b^2+c^2}=30 \end{align}

and the perimeter of $\triangle ABC$

\begin{align} p&=a+b+c=72 . \end{align}

Edit

Expression for the perimeter can be simplified to

\begin{align} p&=\frac{2uv}{3\sqrt{2uv}-2(u+v)} . \end{align}