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In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of ∠BAC. |
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Answer» Consider △ OAB and △ OAC It is given that AB = AC OA is common i.e. OA = OA From the figure we know that OB = OC which is the radii By SSS congruence criterion △ OAB ≅ △ OAC ∠OAB = ∠OAC (c. p. c. t) Therefore, it is proved that O lies on the bisector of ∠BAC. |
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