-->

Areas Of Parallelograms And Triangles

Question
CBSEENMA9002593

In the figure, D and E are points on the base BC of a ∆ABC such that AD = AE and ∠BAD = ∠CAE. Prove that AB = AC.


Solution

Given: D and E are points on the base BC of a ∆ABC such that AD = AE and ∠BAD = ∠CAE.
To Prove: AB = AC
Proof: In ∆ADE,
∵ AD = AE    | Given
∴ ∠ADE = ∠AED    ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABD,
Ext. ∠ADE = ∠BAD + ∠ABD ...(2)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
In ∆AEC,
Ext. ∠AED = ∠CAE + ∠ACE .. .(3)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
From (1), (2) and (3),
∠BAD + ∠ABD = ∠CAE + ∠ACE
⇒ ∠ABD = ∠ACE
| ∵ ∠BAD = ∠CAE (Given)
⇒ ∠ABC = ∠ACB
∴ AB = AC
| Sides opposite to equal angles of a triangle are equal