A) \[\angle BPA=\frac{1}{2}\angle BAC\]
B) \[\angle BAP=\frac{1}{2}\angle BAC\]
C) \[\angle CPA=\frac{1}{2}\angle BAC\]
D) \[\angle BPA=2\angle BAC\]
Correct Answer: B
Solution :
In \[\Delta \Beta \Alpha P\]and \[\Delta CAP\] \[AB=AC\](Sides of isosceles triangle) AP = AP (Common) \[BP=CP\] (Sides of isosceles triangle) \[\therefore \]\[\Delta BAP\cong \Delta CAP\] (By SSS congruency) \[\therefore \]\[\angle BAP=\angle CAP\] (By C. P.C.T) But \[\angle BAP+\angle CAP=\angle BAC\] \[\Rightarrow \]\[2\angle BAP=\angle BAC\Rightarrow \angle BAP=\frac{1}{2}\angle BAC\]
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