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In ∆ ABC, DE || AC, where D and E are two points lying on AB and BC respectively. If AB = 5 cm and AD = 3 cm, then BE : EC is
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- 2 : 3
- 3 : 2
- 5 : 3
- 3 : 5
- 2 : 3
Correct Option: A
We draw a figure triangle ABC in which DE || AC. D and E are two points lying on AB and BC respectively
In ∆ BDE and ∆ ABC,
DE || AC
∴ ∠BDE = ∠BAC
∠BED = ∠BCA
By AA–similarity,
∆BDE ~ ∆BAC
| ⇒ | = | ||
| AD | EC |
| ⇒ | = | ||
| AD | EC |
| ⇒ | = | ||
| 3 | EC |
| ⇒ | = | ||
| EC | 3 |
