Question: The external bisector of $ \angle B $ and $ \angle C $ of $ \Delta ABC $ (where AB and AC extended to E and F, respectively) meet at point P. If $ \angle BAC=100{}^\circ , $ then the measure of $ \angle BPC $ is
Options:
A) $ 50{}^\circ $
B) $ 80{}^\circ $
C) $ 40{}^\circ $
D) $ 100{}^\circ $
Show Answer
Answer:
Correct Answer: C
Solution:
- In $ \Delta ABC $ side AB and AC are produced to E and F, respectively and the external bisector $ \angle EBC $ and $ \angle FCB $ intersect at P.
$ x+y+z=180{}^\circ $
$ y+z=180-x $
$ =180-100=80{}^\circ $
Now, $ 2\angle 1+y=180{}^\circ $
and $ 2\angle 2+z=180{}^\circ $
$ \therefore $ $ 2(\angle 1+\angle 2)=360{}^\circ -(y+z) $
$ =360{}^\circ -80{}^\circ =280{}^\circ $
and $ \angle BPC=180{}^\circ -(\angle 1+\angle 2) $
$ =180{}^\circ -140{}^\circ =40{}^\circ $