In $\angle\text{A}=\angle\text{C},$ AB = 6cm, BP = 15cm, AP = 12cm and CP = 4cm, then find the lengths of PD and CD.
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In $\triangle\text{ABP}$ and $\triangle\text{CDP},$$\angle\text{A}=\angle\text{C}$ [Given]
$\angle1=\angle2$ [Vertically opposite angles]
$\therefore\triangle\text{ABP}\sim\triangle\text{CDP}$ [By AAA similarity criterion]
$\Rightarrow\frac{\text{AB}}{\text{CD}}=\frac{\text{AP}}{\text{CP}}=\frac{\text{BP}}{\text{DP}}$
| $\Rightarrow\frac{6}{\text{y}}=\frac{12}{4}=\frac{15}{\text{x}}$ | $\Rightarrow\frac{15}{\text{x}}=\frac{12}{4}$ |
| $\Rightarrow\frac{6}{\text{y}}=\frac{12}{4}$ | $\Rightarrow\frac{15}{3}=\text{x}$ |
| $\Rightarrow\text{y}=\frac{6}{3}=2\text{cm}$ | $\Rightarrow\text{x}=5\text{cm}$ |
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