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In ∆ ABC, D and E are points on AB and AC respectively such that DE || BC and DE divides the ∆ ABC into two parts of equal areas. Then ratio of AD and BD is
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- 1 : 1
- 1: √2 - 1
- 1: √2
- 1: √2 + 1
Correct Option: B
On the basis of question we draw a figure of triangle ABC in which D and E are points on AB and AC respectively such that DE || BC ,
DE ||BC
∠ADE = ∠ABC
∠AED = ∠ACB
∴ ∆ADE ~ ∆ABC
| Now, | = | ||
| ∆ ADE | 1 |
[DE divides ∆ into two equal parts]
| ⇒ | + 1 = 1 + 1 | |
| ∆ ADE |
| ⇒ | = 2 = | ||
| ∆ ADE | AD² |
| ⇒ | = √2 | |
| AD |
| ⇒ | - 1 = √2 - 1 | |
| AD |
| ⇒ | = √2 - 1 | |
| AD |
| ⇒ | = | ||
| BD | √2 - 1 |
∴ AD : BD = 1 : 2 √2
