91Ó°ÊÓ

The given diagram is:

It is being given that$∠ABCâ‰\dots ∠ACB,\stackrel{¯}{AE}âŠ\yen \stackrel{¯}{EC}$ and $\stackrel{¯}{AD}âŠ\yen \stackrel{¯}{DB}$.

As, $∠ABCâ‰\dots ∠ACB$, therefore by using the theorem $4-2$it can be said that $\stackrel{¯}{AB}â‰\dots \stackrel{¯}{AC}$.

As, $\stackrel{¯}{AE}âŠ\yen \stackrel{¯}{EC}$, therefore it can be said that $m∠AEC=90°$.

As, $\stackrel{¯}{AD}âŠ\yen \stackrel{¯}{DB}$, therefore it can be said that $m∠ADB=90°$.

Therefore, $m∠AEC=m∠ADB=90°$.

Therefore, $∠AECâ‰\dots ∠ADB$.

In the triangles $△ADB$and $△AEC$, the angle that is common is $∠A$.

Therefore, $∠Aâ‰\dots ∠A$by using the reflexive property.

In the triangles$â–³ADB$ and $â–³AEC$, it can be noticed that $∠Aâ‰\dots ∠A,∠AECâ‰\dots ∠ADB$ and$\stackrel{¯}{AB}â‰\dots \stackrel{¯}{AC}$.

Therefore, the triangles $â–³ADB$and $â–³AEC$are the congruent triangles by using the $AAS$method.

The pair of overlapping triangles that are congruent are$â–³ADB$ and $â–³AEC$.

The triangles$â–³ADB$ and$â–³AEC$are congruent by the $AAS$method.