How should we prove this ? boc=90-1/2angle bac

We have m(<CBO) = (1/2) · m(<CBE) = (1/2) · ( x + z );

In the same way, m(<BCO) = (1/2) ·( x + y);

m(<BOC) = 180 - [(1/2) · ( x + z ) + (1/2) ·( x + y)] = 180 - (1/2)· ( x + x + y + z );

But, x + y + z = 180;

Then, m(<BOC) = 180 - (1/2)·( x + 180 );

Finally, m(<BOC) = 90 - (1/2)·x;

So, m(<BOC) = 90 - (1/2)·m(<BAC).

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step-by-step explanation:

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