**Question:**

In figure, two line segments AC and BD intersect each other at the point P such that PA = 6 cm, PB = 3 cm, PC = 2.5 cm, PD = 5 cm, ∠APB = 50° and

∠CDP = 30°. Then, ∠PBA is equal to

(a) 50°

(b) 30°

(c) 60°

(d) 100°

**Solution:**

(d) $\ln \triangle A P B$ and $\triangle C P D$, $\angle A P B=\angle C P D=50^{\circ}$ [vertically opposite angles]

$\frac{A P}{P D}=\frac{6}{5}$ ...(i)

and $\frac{B P}{C P}=\frac{3}{2.5}=\frac{6}{5}$ ...(ii)

From Eqs. (i) and (ii)

$\frac{A P}{P D}=\frac{B P}{C P}$

$\therefore$ $\triangle A P B \sim \triangle D P C$ [by SAS similarity criterion]

$\therefore$ $\angle A=\angle D=30^{\circ}$ [corresponding angles of similar triangles]

In $\triangle A P B, \quad \angle A+\angle B+\angle A P B=180^{\circ} \quad$ [sum of angles of a triangle $=180^{\circ}$ ]

$\Rightarrow \quad 30^{\circ}+\angle B+50^{\circ}=180^{\circ}$

$\therefore \quad \angle B=180^{\circ}-\left(50^{\circ}+30^{\circ}\right)=100^{\circ}$

i.e., $\quad \angle P B A=100^{\circ}$