91Ó°ÊÓ

The figure shows quad. ABCD is a parallelogram.

If two parallel lines are intersected by a transversal then the angle made by a transversal with respect to two lines are called alternate interior angles which are equal in measures.

From the figure, it can be observed that, $\stackrel{¯}{AB}âˆ\yen \stackrel{¯}{CD}$, $\stackrel{¯}{BD}$is a transversal and $∠ABD,∠BDC$, $∠ADB,∠DBC$are alternate interior angles such that

$\begin{array}{l}∠ABDâ‰\dots ∠BDC;\\ ∠ADBâ‰\dots ∠DBC\end{array}$

Consider $\mathrm{Δ}ADB$and $\mathrm{Δ}BCD$, where $∠ABDâ‰\dots ∠BDC$, $∠ADBâ‰\dots ∠DBC$and $\stackrel{¯}{BD}$is common side in both the triangles therefore by ASA postulate, $\mathrm{Δ}ADBâ‰\dots \mathrm{Δ}BCD$.

As $\mathrm{Δ}ADBâ‰\dots \mathrm{Δ}BCD$, the angles of congruent triangles are congruent, that is,$∠Aâ‰\dots ∠C$.

From the figure, it can be observed that, $\stackrel{¯}{AD}âˆ\yen \stackrel{¯}{BC}$, $\stackrel{¯}{AC}$is a transversal and $∠DCA,∠CAB$, $∠DAC,∠ACB$are alternate interior angles such that

$\begin{array}{l}∠DCAâ‰\dots ∠CAB;\\ ∠DACâ‰\dots ∠ACB\end{array}$

Consider $\mathrm{Δ}ADC$and $\mathrm{Δ}ABC$, where $∠DCAâ‰\dots ∠CAB$, $∠DACâ‰\dots ∠ACB$and $\stackrel{¯}{AC}$is common side in both the triangles therefore by ASA postulate, $\mathrm{Δ}ADCâ‰\dots \mathrm{Δ}ABC$.

As $\mathrm{Δ}ADCâ‰\dots \mathrm{Δ}ABC$, the angles of congruent triangles are congruent, that is, $∠Bâ‰\dots ∠D$.

As $∠A,∠C$and $∠B,∠D$are opposite angles of a parallelogram.

From $∠Aâ‰\dots ∠C$and $∠Bâ‰\dots ∠D$it can be concluded that the opposite angles of a parallelogram are congruent.