Q. 18

# In ΔABC, sides AB and AC are produced to D and E respectively. BO and CO are the bisectors of ∠CBD and ∠BCE respectively. Then, prove that

Here BO, CO are the angle bisectors of DBC &ECB intersect each other at O.

∴∠1 = 2 and 3 = 4

Side AB and AC of ΔABC are produced to D and E respectively.

Exterior of DBC = A + C ………… (1)

And Exterior of ECB = A + B ………… (2)

Adding (1) and (2) we get

DBC + ECB = 2 A + B + C.

22 + 23 = A + 180°

2 + 3 = (1 /2)A + 90° ………… (3)

But in a ΔBOC = 2 + 3 + BOC = 180° ………… ( 4)

From eq (3) and (4) we get

(1 /2)A + 90° + BOC = 180°

BOC = 90° – (1 /2)A

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