Bisector of ∠A intersects circumcircle of ∆ ABC at D. If m ∠BCD-50, then find m∠BAC.
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Given- ΔABC has been inscribed in a circle. AE, the bisector of ∠BAC, meets BC at D and the arc BEC at E. ∠ECD=30
o
when EC is joined.
To find out- ∠BAC=? Solution- We join BE. Now, BE, the chord of the given circle, subtends ∠BAE&∠BCE to the circumference of the given circle at A & C respectively. So ∠BAE=∠BCE.......(i) (since angles, subtended by a chord of a circle to the circumference of the same circle at different points, are equal.)
But ∠BAE=∠EAC........(ii) since AE is the bisector of \angle BAC.
∴ From (i) & (ii) ∠BCE=∠BAE=∠EAC=30
o
.
∴∠BAC=∠BAE+∠EAC=30
o
+30
o
=60
o
.
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