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Question
In the following figure, D and E are points on side BC of a ∆ABC such that BD = CE and AD = AE. Show that ∆ABD ≅ ∆ACE.
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Solution
Given: D and E are the points on side BC of a ∆ABC such that BD = CE and AD = AE.
To show: ∆ABD ≅ ∆ACE
Proof: We have, AD = AE ...[Given]
⇒ ∠ADE = ∠AED ...(i) [Since, angles opposite to equal sides are equal]
We have, ∠ADB + ∠ADE = 180° ...[Linear pair axiom]
⇒ ∠ADB = 180° – ∠ADE
= 180° – ∠AED ...[From equation (i)]
In ∆ABD and ∆ACE,
∠ADB = ∠AEC ...[∵ ∠AEC + ∠AED = 180°, linear pair axiom]
BD = CE ...[Given]
And AD = AE ...[Given]
∴ ∆ABD ≅ ∆ACE ...[By SAS congruence rule]
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