In the figure,two line segments AC and BD int | vedclass

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(C) In $	riangle APB$ and $	riangle DPC$,we have:$\angle APB = \angle DPC = 50^{\circ}$ [Vertically opposite angles]Now,consider the ratios of the s

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$100$

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(C) In $	riangle APB$ and $	riangle DPC$,we have:$\angle APB = \angle DPC = 50^{\circ}$ [Vertically opposite angles]Now,consider the ratios of the sides including the equal angles:$\frac{PA}{PD} = \frac{6}{5} = 1.2$$\frac{PB}{PC} = \frac{3}{2.5} = \frac{30}{25} = 1.2$Since $\frac{PA}{PD} = \frac{PB}{PC}$ and the included angles are equal,by the $SAS$ similarity criterion:$	riangle APB \sim 	riangle DPC$Since the triangles are similar,their corresponding angles are equal:$\angle PAB = \angle PDC = 30^{\circ}$In $	riangle APB$,the sum of angles is $180^{\circ}$:$\angle PAB + \angle APB + \angle PBA = 180^{\circ}$$30^{\circ} + 50^{\circ} + \angle PBA = 180^{\circ}$$80^{\circ} + \angle PBA = 180^{\circ}$$\angle PBA = 180^{\circ} - 80^{\circ} = 100^{\circ}$

Part $I$	Part $II$
$1.$ In $\Delta ABC$,$\angle B$ is a right angle and $\overline{BM}$ is a median.	$a. AB^2 + BC^2 = 2(BD^2 + CD^2)$
$2.$ In $\Delta ABC$,$\angle A$ is a right angle and $\overline{AD}$ is an altitude.	$b. BC = \frac{1}{2} AB$
$3.$ In $\Delta ABC$,$m\angle C = 90^\circ$ and $m\angle A = 30^\circ$.	$c. AC^2 = CD \cdot BC$
$4.$ In $\Delta ABC$,$\overline{BD}$ is a median.	$d. BM = \frac{1}{2} AC$

In the figure,two line segments $AC$ and $BD$ intersect each other at the point $P$ such that $PA = 6 \, cm$,$PB = 3 \, cm$,$PC = 2.5 \, cm$,$PD = 5 \, cm$,$\angle APB = 50^{\circ}$ and $\angle CDP = 30^{\circ}$. Then,$\angle PBA$ is equal to: (in $^{\circ}$)

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