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In fig, AB = AC, D is a point on AC and E on AB such that AD = ED = EC = BC. Then ∠A : ∠B :
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- 1 : 2
- 2 : 1
- 3 : 1
- 1 : 3
- None of these
Correct Option: D
As per given figure , we can see that
In ΔBCE, BC = EC, ∴ ∠B = ∠BEC
In ΔCDE, ED = EC, ∴ ∠ECD = ∠EDC
and In ΔADE, AD = ED, ∴ ∠AED = ∠A
Now ∠B = ∠BEC = ∠A + ∠ECD = ∠A + ∠EDC = ∠A + ∠EAD + ∠AED
{ ∴ ∠EDC = ∠EAD + ∠AED }
=∠A + ∠A + ∠A = 3∠A { By ASA property }
| ∴ | ∠B | = | ∠3 |
| ∠A | 1 |
⇒ ∠A : ∠B = 1 : 3.
