Question

BC is a chord with centre O. A is a point on an arc BC as shown in the figure. Prove that:

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(i) $\angle BAC+\angle OBC={90}^{\circ }$, if A is the point on the major arc

(ii) $\angle BAC-\angle OBC={90}^{\circ }$, if A is the point on the minor arc.

Answer 1

(i) We observe that the minor arc BC makes $\angle BOC=2\angle BAC$ at the centre

Let $\angle BAC=x$

$\therefore z=2x$ ...... (1)

In $△OBC$, we have

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

$y+z+y={180}^{\circ }$

$2x+2y={180}^{\circ }$ [From (1)]

$x+y={90}^{\circ }$

$\angle BAC+\angle OBC={90}^{\circ }$


(ii) We observe that the major arc BC subtends $\angle BOC=z$ and $\angle O=t$, at the centre and $\angle BAC=x$, at a point on the circumference.

$\therefore z=2x$

In $\Delta OBC$, we have

$\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$

$y+y+t={180}^{\circ }$

$t={180}^{\circ }-2y$

Now, $z={360}^{\circ }-t$

$z={360}^{\circ }-{180}^{\circ }+2y$

$2x={180}^{\circ }+2y[z=2x]$

$x-y={90}^{\circ }$

$\angle BAC-\angle OBC={90}^{\circ }$