In Figure, D and E are points on side BC of a ΔABC such that BD = CE and AD = AE. Show that ΔABD ≅ ΔACE.
Given: In ΔABC, BD = CE and AD = AE.
In ΔADE,
AD = AE
⇒ ∠ADE = ∠AED …(1) (opposite angles to equal sides are equal)
Now, ∠ADE + ∠ADB = 180° (linear pair)
∠ADB = 180° - ∠ADE ..(2)
Also, ∠AED + ∠AEC = 180° (linear pair)
∠AEC = 180° - ∠AED
∠AEC = 180° - ∠ADE .. (3) (∵∠ADE = ∠AED)
From (2) and (3)
∠ADB = ∠AEC ..(4)
Now, In ΔADB and ΔAEC,
AD = AE (given)
BD = EC (given)
∠ADB = ∠AEC (from (4)
Hence, ΔABD ≅ΔACE (by SAS)