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Q12.
ExpertverifiedWrite proof in twocolumn form.
Given: $\overline{QR}$, $\overline{ST}$ bisect each other. Prove: $\angle XRT>m\angle S$.
Statement  Proof 
1. $QV=VR$, $SV=VT$  Given 
2. $\angle QVS=\angle RVT$  Vertically opposite angles 
3. $\u25b3QSV\cong \u25b3VRT$  SAS 
4. $\angle T=\angle S$  CPCT 
5. $\angle XRT=\angle RVT+\angle S$ 

6. $m\angle XRT>m\angle S$  Property of Inequality If $a=b+c$ and $c>0$ then $a>c$. 
Two triangles $QSV$ and $VRT$ as shown below are the given triangles.
If two sides and one angle equal of two triangles are equal then from SAS corresponding triangles will be congruent triangles.
Since QR and ST bisect each other, therefore $QV=VR$, $SV=VT$.
The vertically opposite angles are always equal.
Therefore, $\angle QVS=\angle RVT$. Then by SAS, $\u25b3QSV\cong \u25b3VRT$.
Since $\u25b3QSV\cong \u25b3VRT$ then by CPCT $\angle T=\angle S$.
Use exterior angle property as follows:
$\begin{array}{l}\angle XRT=\angle RVT+\angle T\\ \angle XRT=\angle RVT+\angle S\end{array}$
If $a=b+c$ and $c>0$ then $a>c$.
Since $\angle XRT=\angle RVT+\angle S$ therefore $m\angle XRT>m\angle S$
Hence proved.
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