Question
In $\triangle ABC , \angle B =35^{\circ}, \angle C =65^{\circ}$ and the bisector of $\angle BAC$ meets$ BC$ in $P$ . Arrange $AP , BP$ and $CP$ in descending order.
✓
Answer
In angle $A+B+C=180 A+35+65=180 A=180-100 A=80$
So $\angle BAP$ and $\angle CAP =\frac{80}{2}=40$
we know that side opposite to the greater angle is longer $\ln \triangle BAP$
we know that side opposite to the greater angle is longer so $BP > AP$
In $\triangle CAP ACP (65)> CAP (40)$ so $AP > CP$ so $BP > AP > CP$
So $\angle BAP$ and $\angle CAP =\frac{80}{2}=40$
we know that side opposite to the greater angle is longer $\ln \triangle BAP$
we know that side opposite to the greater angle is longer so $BP > AP$
In $\triangle CAP ACP (65)> CAP (40)$ so $AP > CP$ so $BP > AP > CP$
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