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In Fig. 6.38, the sides AB and AC of AB...

In Fig. 6.38, the sides AB and AC of ABC are produced to points E and D respectively. If bisectors BO and CO of CBE and BCD respectively meet at point O, then prove that `/_B O C=90^(@)-1/2/_B A C`.

Text Solution

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`∠CBE = 180 - ∠ABC`
`∠CBO = 1/2 ∠CBE` (BO is the bisector of ∠CBE)
`∠CBO = 1/2 ( 180 - ∠ABC) `
` 1/2 x 180 = 90 ` `∠CBO = 90 - 1/2 ∠ABC` .............(1)
` 1/2 x ∠ABC = 1/2∠ABC` ...
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