Find the pair of similar triangles among the given pairs.


Find the pair of similar triangles among the given pairs. State the similarity criterion and write the similarity relation in symbolic form.


We have:

$\angle B A C=\angle P Q R=50^{\circ}$

$\angle A B C=\angle Q P R=60^{\circ}$


$\angle A C B=\angle P R Q=70^{\circ}$

Therefore, by AAA similarity theorem, $\triangle A B C \sim Q P R$

We have:

$\frac{A B}{D F}=\frac{3}{6}=\frac{1}{2}$ and $\frac{B C}{D E}=\frac{4.5}{9}=\frac{1}{2}$

But, $\angle A B C \neq \angle E D F$ (Included angles are not equal)

Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

We have:

$\frac{C A}{Q R}=\frac{8}{6}=\frac{4}{3}$ and $\frac{C B}{P Q}=\frac{6}{4.5}=\frac{4}{3}$

$\Rightarrow \frac{C A}{Q R}=\frac{C B}{P Q}$

Also, $\angle A C B=\angle P Q R=80^{\circ}$

Therefore, by SAS similarity theorem, $\triangle A C B \sim \triangle R Q P$.

 We have

$\frac{D E}{Q R}=\frac{2.5}{5}=\frac{1}{2}$

$\frac{E F}{P Q}=\frac{2}{4}=\frac{1}{2}$

$\frac{D F}{P R}=\frac{3}{6}=\frac{1}{2}$

$\Rightarrow \frac{D E}{Q R}=\frac{E F}{P Q}=\frac{D F}{P R}$

Therefore, by SSS similarity theorem, $\triangle F E D \sim \triangle P Q R$


In $\triangle \mathrm{ABC}$

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \quad$ (Angle Sum Property)

$\Rightarrow 80^{\circ}+\angle B+70^{\circ}=180^{\circ}$

$\Rightarrow \angle \mathrm{B}=30^{\circ}$

$\angle \mathrm{A}=\angle \mathrm{M}$ and $\angle \mathrm{B}=\angle \mathrm{N}$

Therefore, by AA similarity theorem, $\triangle \mathrm{ABC} \sim \triangle \mathrm{MNR}$

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