$\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$.
Question.
$\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$. Show that $\triangle \mathrm{RPQ} \sim$ $\triangle \mathrm{RTS}$.
$\mathrm{S}$ and $\mathrm{T}$ are points on sides $\mathrm{PR}$ and $\mathrm{QR}$ of $\triangle \mathrm{PQR}$ such that $\angle \mathrm{P}=\angle \mathrm{RTS}$. Show that $\triangle \mathrm{RPQ} \sim$ $\triangle \mathrm{RTS}$.
Solution:
In figure, We have RPQ and RTS in which
$\angle \mathrm{RPQ}=\angle \mathrm{RTS}$ (Given)
$\angle \mathrm{PRQ}=\angle \mathrm{SRT}(\mathrm{Each}=\angle \mathrm{R})$
Then by AA similarity criterion, we have
$\Delta \mathrm{RPQ} \sim \Delta \mathrm{RTS}$
In figure, We have RPQ and RTS in which
$\angle \mathrm{RPQ}=\angle \mathrm{RTS}$ (Given)
$\angle \mathrm{PRQ}=\angle \mathrm{SRT}(\mathrm{Each}=\angle \mathrm{R})$
Then by AA similarity criterion, we have
$\Delta \mathrm{RPQ} \sim \Delta \mathrm{RTS}$
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