91Ó°ÊÓ

The given diagram is:

It is being given that$\stackrel{¯}{PQ}âŠ\yen \stackrel{¯}{QR}$, $\stackrel{¯}{PS}âŠ\yen \stackrel{¯}{SR}$and$∠1â‰\dots ∠4$.

As, $\stackrel{¯}{PQ}âŠ\yen \stackrel{¯}{QR}$, therefore $∠PQR=90°$.

As, role="math" localid="1646591652918" $\stackrel{¯}{PS}âŠ\yen \stackrel{¯}{SR}$, therefore $∠PSR=90°$.

From the given diagram it can be noticed that:

$∠PQS+∠RQS=∠PQR$

Therefore, it can be obtained that:

$\begin{array}{c}∠PQS+∠RQS=∠PQR\\ ∠1+∠2=90°\end{array}$

As, $∠1+∠2=90°$, therefore∠1 and∠2are complementary angles.

From the given diagram it can be noticed that:

$∠PSQ+∠RSQ=∠PSR$

Therefore, it can be obtained that:

$\begin{array}{c}∠PSQ+∠RSQ=∠PSR\\ ∠4+∠5=90°\end{array}$

As, $∠4+∠5=90°$, therefore∠4 and∠5are complementary angles.

As,$∠1+∠2=90°$ and $∠4+∠5=90°$, therefore by using the substitutive property it can be obtained that:

$∠1+∠2=∠4+∠5$

As, $∠1â‰\dots ∠4$, therefore it can be obtained that:

$\begin{array}{c}∠1+∠2=∠4+∠5\\ ∠1+∠2=∠1+∠5\\ ∠2=∠5\end{array}$

Therefore, $∠2â‰\dots ∠5$.

Hence proved.

The complete proof table is:

The given diagram is:

It has been obtained that$∠2â‰\dots ∠5$.

From the diagram, it can be noticed that angles ∠2 and ∠3 forms the linear pair.

Therefore, $∠2+∠3=180°$.

From the diagram, it can be noticed that angles ∠5 and ∠6 form the linear pair.

Therefore, $∠5+∠6=180°$.

As,$∠2+∠3=180°$ and $∠5+∠6=180°$, therefore by using the substitutive property it can be obtained that:

$∠2+∠3=∠5+∠6$

As,$∠2â‰\dots ∠5$, therefore it can be obtained that:

$\begin{array}{c}∠2+∠3=∠5+∠6\\ ∠2+∠3=∠2+∠6\\ ∠3=∠6\end{array}$

Therefore, $∠3â‰\dots ∠6$.

Hence proved.