Answer :


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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