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Question
In the figure, point D and E are on side BC of ΔABC, such that BD = CE and AD = AE. Show that ΔABD ≅ ΔACE.
Solution
In ∆ADE,
seg AD = seg AE ...[Given]
∴ ∠ADE = ∠AED ...(1) (Isosceles triangle theorem)
Now,
∠ADE + ∠ADB = 180° ...(2) [Angles in a linear pair]
∠AED + ∠AEC = 180° ...(3) [Angles in a linear pair]
∴ ∠ADE + ∠ADB = ∠AED + ∠AEC ...[From (2) and (3)]
∴ ∠ADE + ∠ADB = ∠ADE + ∠AEC ...[From (1)]
⇒ ∠ADB = ∠AEC ...(4)
In ∆ABD and ∆ACE
∠ADE = ∠AEC ...[From (4)]
seg BD ≅ seg CE ...[Given]
seg AD ≅ seg AE ...[Given]
By SAS test of congruency
∴ ∆ABD ≅ ∆ACE ...[SAS test]
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