Triangle problem, $AC=3$ , $AB=5$ ,..., then $PH=?$

Question

In $\triangle ABC$ :
$AC=3$ , $AB=5$ , $\angle ACB= 90 ^ \circ$,
$P$ is a point inside $\triangle ABC$ such that $PA+BC=PB+AC=PC+AB$,
$H$ is a point on the line $AB$ such that $\angle PHB=90^\circ$
then, $PH=?$

My Approach:
In Right-angled $\triangle ABC$
$$AB^2=AC^2 + BC^2 \implies BC=4$$ Let $PC=x$
then, $$PA+4=PB+3=PC+5 $$ $$[\because PA+BC=PB+AC=PC+AB]$$ $$\implies PA=x+1 \, , \, PB= x+2$$ so,

My thoughts: First we will find the value of $x$ by equating area of triangle $ABC$,
i.e, $$Ar[\triangle ABC]=Ar[\triangle APB]+Ar[\triangle BPC]+Ar[\triangle APC] \dots (i)$$ $Ar[\triangle ABC]$ can be simply found as: $\frac{1}{2}*3*4=6$ sq. units
For $Ar[\triangle APB]$ , we will use Heron's formula as follows: $$s=\frac{1}{2} * [(x+1)+(x+2)+5] = x+4 $$ $$\therefore Ar[\triangle APB]=\sqrt{(x+4)(x-1)(2)(3)}=\sqrt{6(x+4)(x-1)}$$ Similarly, $Ar[\triangle BPC]=\sqrt{3(x+3)(x-1)}$
Similarly, $Ar[\triangle APC]=\sqrt{2(x+2)(x-1)}$
So, from eq.(i) , we get: $$6=\sqrt{6(x+4)(x-1)} \, + \, \sqrt{3(x+3)(x-1)} \, + \, \sqrt{2(x+2)(x-1)} \dots (ii)$$ and then we will find $PH$ by equating area of triangle $APB$,
i.e, $$\frac{1}{2}*5*PH=\sqrt{6(x+4)(x-1)} \dots (iii)$$ So, how to solve eq.(ii) to find the value of $x$ ? Please help...

Answer 1

Hint: Let $x$ the perpendicular on $AC$,$y$ the perpendicular on $BC$ then we get the following equations: $$3\cdot 4=3x+4y+5PH$$ Now let $$PA=t-BC,PB=t-AC,PC=t-AB$$ then we obtain

$$(t-5)^2=x^2+y^2$$

$$(t-3)^2=(4-x)^2+y^2$$

$$(t-4)^2=x^2+(3-y)^2$$ Can you finish? I got $$PH=\frac{132}{115},x=\frac{20}{23},y=\frac{21}{23},t=\frac{144}{23}$$

Answer 2

Continuing from your labels!

Drop perpendiculars $y$ and $z$ from the point $P$ to the sides $AC$ and $BC$, respectively, as shown:

$\hspace{1cm}$

Make up the system of equations: $$\begin{cases}y^2+z^2=x^2\\ y^2+(3-z)^2=(x+1)^2\\ z^2+(4-y)^2=(x+2)^2\\ \frac12\cdot (3y+4z+5h)=\frac12\cdot 3\cdot 4 \ \text{(area of triangle $ABC$)}\end{cases}$$

Can you solve it?

WolframAlpha answer:

$h=\frac{132}{115}, x=\frac{29}{23},y=\frac{20}{23},z=\frac{21}{23}.$