bisectors in a right triangle

Question

Given right triangle $ABC$, $AC=8$, $BC=6$ and $\angle C = \frac{\pi}{2}$.

$BP$ and $CR$ – bisectors of triangle $ABC$, intersecting at the point $L$.

Need to find:

1) periods in which these bisectors divide point $L$.

2) the area of the quadrilateral $APLR$.

How to solve these problems? Unfortunately so far, no ideas. Where to start?

Answer 1

By Pythagoras theorem we obtain $AB=10$.

Now, $AP:BC=10:6=5:3$, which gives $AP=5$ and $PC=3$.

Also, $AR:RB=8:6=4:3,$ which gives $RB=\frac{30}{7}$.

Thus, $CL:LR=6:\frac{30}{7}=7:5$ and $PL:LB=3:6=1:2$ and 1) is done!

$S_{\Delta ABC}=\frac{8\cdot6}{2}=24$.

Hence, $S_{\Delta APB}=\frac{5}{8}\cdot24=15$.

Now, $S_{\Delta CBR}=\frac{3}{7}\cdot24=\frac{72}{7}$.

Thus, $S_{\Delta LBR}=\frac{5}{12}\cdot\frac{72}{7}=\frac{30}{7}$.

Id est, $S_{APLR}=15-\frac{30}{7}=\frac{75}{7}$ and we are done!

Answer 2

\begin{align} A&=(6,8),\quad B=(0,0),\quad C=(0,8) ,\\ a&=|BC|=6 ,\quad b=|AC|=8 ,\quad c=|AB|=10. \end{align}

$L$ is incenter of $\triangle ABC$, $P$ and $Q$ are the feet points of the bisectors, \begin{align} L&=\frac{aA+bB+cC}{a+b+c}=(4,\,2) ,\\ P&=\frac{aA+cC}{a+c}=(6,\,3) ,\\ Q&=\frac{aA+bB}{a+b}=(\tfrac{18}7,\,\tfrac{24}7) ,\\ |BP|&=|B-P|=3\,\sqrt5 ,\quad |LP|=|L-P|=\sqrt5 ,\\ |CQ|&=|C-Q|=\tfrac{24}7\sqrt2 ,\quad |LQ|=|L-Q|= \tfrac{10}7\sqrt2 . \end{align}

By the shoelace formula for the area of a simple polygon whose vertices are described by their Cartesian coordinates,

\begin{align} S_{AQLP} &=\tfrac12( A_x\,Q_y+Q_x\,L_y+L_x\,P_y+P_x\,A_y-(A_y\,Q_x+Q_y\,L_x+L_y\,P_x+P_y\,A_x) )=\tfrac{75}{7} . \end{align}