91Ó°ÊÓ

If two sides of a triangle are congruent then the angles opposite to those sides are congruent.

Consider$\mathrm{Δ}POQ$ in which$\stackrel{¯}{PO}â‰\dots \stackrel{¯}{QO}$ then by isosceles triangle theorem, $∠1â‰\dots ∠2$, that is, $m∠2=m∠1=40$.

Consider$\mathrm{Δ}POQ$ then by an angle sum theorem,

$$\begin{gathered} m∠1+m∠2+m∠7=180 \\ 40+40+m∠7=180 \\ 80+m∠7=180 \\ m∠7=100 \end{gathered}$$

From the given figure, it can be observed that, $∠7$and $∠O$are vertically opposite angles therefore they are equal in measure, that is,

$m∠O=m∠7=100$

Consider$\mathrm{Δ}SOR$ in which$\stackrel{¯}{RO}â‰\dots \stackrel{¯}{SO}$ then by isosceles triangle theorem, $∠5â‰\dots ∠6$, that is, $m∠5=m∠6$.

Consider$\mathrm{Δ}SOR$ then by an angle sum theorem,

$$\begin{gathered} m∠5+m∠6+m∠O=180 \\ m∠5+m∠5+m∠7=180 \\ 2m∠5+100=180 \\ 2m∠5=80 \\ m∠5=40 \end{gathered}$$

Therefore, $m∠6=m∠5=40$.

As $m∠1=m∠6$implies that $∠1â‰\dots ∠6$and are alternate interior angles where $\stackrel{¯}{PR}$is the transversal to the line segments $\stackrel{¯}{PQ}$and $\stackrel{¯}{SR}$, then by the Converse of Alternate Interior Angle Theorem, $\stackrel{¯}{PQ}âˆ\yen \stackrel{¯}{SR}$.

Therefore, the measures of$∠2,∠7,∠5$ and$∠6$ are$40,100,40$ and$40$ respectively and $\stackrel{¯}{PQ}$must be parallel to $\stackrel{¯}{SR}$.

If two sides of a triangle are congruent then the angles opposite to those sides are congruent.

Consider$\mathrm{Δ}POQ$ in which$\stackrel{¯}{PO}â‰\dots \stackrel{¯}{QO}$ then by isosceles triangle theorem, $∠1â‰\dots ∠2$, that is, $m∠2=m∠1=k$.

Consider$\mathrm{Δ}POQ$ then by an angle sum theorem,

$$\begin{gathered} m∠1+m∠2+m∠7=180 \\ k+k+m∠7=180 \\ 2k+m∠7=180 \\ m∠7=180−2k \end{gathered}$$

From the given figure, it can be observed that, $∠7$and $∠O$are vertically opposite angles therefore they are equal in measure, that is,

$m∠O=m∠7=180−2k$

Consider$\mathrm{Δ}SOR$ in which$\stackrel{¯}{RO}â‰\dots \stackrel{¯}{SO}$ then by isosceles triangle theorem, $∠5â‰\dots ∠6$, that is, $m∠5=m∠6$.

Consider $\mathrm{Δ}SOR$then by an angle sum theorem,

$$\begin{gathered} m∠5+m∠6+m∠O=180 \\ m∠5+m∠5+m∠7=180 \\ 2m∠5+180−2k=180 \\ 2m∠5=2k \\ m∠5=k \end{gathered}$$

Therefore, $m∠6=m∠5=k$.

As $m∠1=m∠6$implies that $∠1â‰\dots ∠6$and are alternate interior angles where $\stackrel{¯}{PR}$is the transversal to the line segments $\stackrel{¯}{PQ}$and $\stackrel{¯}{SR}$, then by the Converse of Alternate Interior Angle Theorem, $\stackrel{¯}{PQ}âˆ\yen \stackrel{¯}{SR}$.

Therefore, the measures of$∠2,∠7,∠5$ and$∠6$ are$k,180−2k,k$ and$k$ respectively and $\stackrel{¯}{PQ}$must be parallel to $\stackrel{¯}{SR}$.