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…
Best Answer
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 $$PH=\frac{132}{115},x=\frac{20}{23},y=\frac{21}{23},t=\frac{144}{23}$$