BC = AB cos B
a = c cos B,
and
b cos C = b cos 900 = 0 Thus, we may write
a = 0 + c cos B
= b cos C + c cos B (3)
From (1), (2) and (3) We assert that for any triangle ABC,
a = b cos C + c cos B
By case (i), r ⇒ q is proved