Question

In the figure, point D and E are on side BC of $∆$ABC, such that BD = CE and AD = AE.

Show that $∆$ ABD$\cong$ $∆$ACE.

Answer 1

​In △ADE, AD = AE
∴ ∠ADE = ∠AED (Angles opposite to equal sides)
Now,
∠ADE + ∠ADB = 180^∘ ...(1)
∠AED + ∠AEC = 180^∘ ...(2)
Subtracting (2) from (1), we get
∠ADB − ∠AEC = 0
⇒ ∠ADB = ∠AEC ...(3)
In △ABD and △ACE
∠ADE = ∠AEC [From (3)]
BD = CE (Given)
AD = AE (Given)
By SAS test of congruency
△ABD ≅ △ACE