Question
If $\angle1=\angle2$ and $\triangle\text{NSQ}\cong\triangle\text{MTR},$ then prove that $\triangle\text{PTS}\sim\triangle\text{PRQ}.$
✓
Answer
Given $\triangle\text{NSQ}\cong\triangle\text{MTR},$
and $\angle1=\angle2$
To prove $\triangle\text{PTS}\sim\triangle\text{PRQ}$
Proof since, $\triangle\text{NSQ}\cong\triangle\text{MTR}$
So, $\text{SQ}=\text{TR}\ ......(\text{i})$
Also, $\angle1=\angle2\Rightarrow\text{PT}=\text{PS}\ ......(\text{ii})$
[Since, sides opposite to equal angles are also equal]
From Eq. (i) and (ii),
$\frac{\text{PS}}{\text{SQ}}=\frac{\text{PT}}{\text{TR}}$
$\Rightarrow\text{ST}\parallel\text{QR}$ [by convense of basic proportionality therom]
$\therefore\angle1=\angle\text{PQR}$
and $\therefore\angle2=\angle\text{PRQ}$ [common angles]
In $\triangle\text{PTS}$ and $\triangle\text{PRQ},$
$\angle{\text{P}}=\angle{\text{P}}$
$\angle1=\angle{\text{PQR}}$
$\angle2=\angle{\text{PRQ}}$
$\therefore\triangle{\text{PTS}}\sim\triangle{\text{PRQ}}$ [by AAA similarity criterion]
Hence proved.
and $\angle1=\angle2$
To prove $\triangle\text{PTS}\sim\triangle\text{PRQ}$
Proof since, $\triangle\text{NSQ}\cong\triangle\text{MTR}$
So, $\text{SQ}=\text{TR}\ ......(\text{i})$
Also, $\angle1=\angle2\Rightarrow\text{PT}=\text{PS}\ ......(\text{ii})$
[Since, sides opposite to equal angles are also equal]
From Eq. (i) and (ii),
$\frac{\text{PS}}{\text{SQ}}=\frac{\text{PT}}{\text{TR}}$
$\Rightarrow\text{ST}\parallel\text{QR}$ [by convense of basic proportionality therom]
$\therefore\angle1=\angle\text{PQR}$
and $\therefore\angle2=\angle\text{PRQ}$ [common angles]
In $\triangle\text{PTS}$ and $\triangle\text{PRQ},$
$\angle{\text{P}}=\angle{\text{P}}$
$\angle1=\angle{\text{PQR}}$
$\angle2=\angle{\text{PRQ}}$
$\therefore\triangle{\text{PTS}}\sim\triangle{\text{PRQ}}$ [by AAA similarity criterion]
Hence proved.
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