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In the figure, BD and CD are angle bisectors of ∠ ABC and ∠ ACE, respectively. Then ∠ BDC is equal to :
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- ∠BAC
- 2∠BAC
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1 ∠BAC 2 -
1 ∠BAC 3 - None of these
Correct Option: C
From above given figure , we have
In ΔABC, ∠ACE = ∠ABC + ∠BAC
Similarly in ΔBCD, ∠BDC = ∠DCE − ∠DBC [Ext. angle prop. of a Δ]
| But ∠DCE = | 1 | ∠ACE and |
| 2 |
| ⇒ | 1 | ∠DBC = | 1 | ∠ABC |
| 2 | 2 |
Now ∠BDC = ∠DCE - ∠DBC
| = | 1 | (∠ACE - ∠ABC) |
| 2 |
| = | 1 | (∠ACE + ∠ABC - ∠ACE) |
| 2 |
| ∴ ∠BDC = | 1 | ∠BAC |
| 2 |
