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ABC is a Δ in which AB = AC and D is a point on AC such that BC2 = AC × CD. Then :
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- BD = DC
- BD = BC
- BD = AB
- BD = AD
- None of these
Correct Option: B
In ΔABC , Given , AB = AC
Since BC2 = AC × CD
⇒ BC x BC = AC × CD
| ∴ | BC | = | AC | and ∠ ACB =∠ BCD |
| CD | BC |
∴ ΔABC ∼ ΔBDC (SAS Similarity)
| AB | = | BC | = | AC |
| BD | DC | BC |
But AB = AC
| ∴ | 1 | = | 1 | ⇒ BD = BC |
| BD | BC |
