Question

If the bisectors of angle abc and angle acb of triangle abc meet at o prove that angle boc is equal to 90 degree + half angle a

Answer 1

$in \: abc \: 2x + 2y + a = 180 \\ a = 180 - 2(x + y) \\ 2(x + y) = 180 - a \\ x + y = 90 - a \div 2 \\ in \: obc \: boc = 180 - (90 - a \div 2) \\ \: \: \: \: = 90 + a \div 2$