Question

In the adjacent figure, the bisector of $\angle A$ and $\angle B$ meet in a point $P$. If $\angle C={100}^o$ and $\angle D={60}^o$, find the measure of $\angle APB$.

Answer 1

$AB\parallel CD$

$\angle B+\angle C=180$

$\angle B=180-100=80$

$\angle PBC+\angle PBA=80$

$2\angle PBC=80$

$\angle PBC=40$

$\angle A+\angle D=180$

$\angle A=180-60=120$

$2\angle PAB=120$

$\angle PAB=120/2=60$

In $△PAB,\angle PAB+\angle ABP+\angle BPA=180$

$60+\angle APB+40=180$

$\angle APB=180-100$

$\angle APB=80$

$\therefore \angle APB=80$