+1839 Point $P$ is on side $\overline{AC}$ of triangle $ABC$ such that $\angle APB =\angle PBA$, and $\angle ABC - \angle ACB = 24^\circ$. Find $\angle PBC$ in degrees.
+195 Since ∠APB=∠PBA, triangles ABP and PBA are isosceles. This means that ∠ABP=∠BAP=21(180∘−∠APB).
Similarly, since ∠BPC=∠PBC, triangles BPC and PBC are isosceles. This means that ∠BPC=∠PBC=21(180∘−∠BPC).
Summing the angle measures in △ABC gives $ \angle ABC + \angle ACB + \angle BAC = 180^\circ$. Since ∠APB=∠PBA and ∠BPC=∠PBC, we can rewrite this as [ \angle ABP + \angle BAP + \angle BPC + \angle PBC + \angle BAC = 180^\circ. ] Substituting what we found for each angle measure, we get [ \frac{1}{2} (180^\circ - \angle APB) + \frac{1}{2} (180^\circ - \angle APB) + \frac{1}{2} (180^\circ - \angle BPC) + \frac{1}{2} (180^\circ - \angle BPC) + \angle BAC = 180^\circ. ] Simplifying the left side gives [ 2 \cdot 180^\circ - (\angle APB + \angle BPC) + \angle BAC = 180^\circ. ] We are given that ∠ABC−∠ACB=24∘, which is the same as ∠BAP−∠BPC=24∘ since ∠APB=∠BAP and ∠BPC=∠PBC. Substituting again, we get [ 2 \cdot 180^\circ - (\angle BAP + \angle BPC) + \angle BAC = 180^\circ \Rightarrow 2 \cdot 180^\circ - (180^\circ + 24^\circ) + \angle BAC = 180^\circ. ] Solving for ∠BAC gives ∠BAC=36∘.
Since the angles in a triangle sum to 180∘, we can find ∠PBC as follows: [ \angle PBC = 180^\circ - \angle BPC - \angle BAC = 180^\circ - \frac{1}{2} (180^\circ - \angle BPC) - 36^\circ. ] Substituting ∠BAP−∠BPC=24∘ again, we get [ \angle PBC = 180^\circ - \frac{1}{2} (180^\circ - (\angle BAP + 24^\circ)) - 36^\circ = 180^\circ - 90^\circ - 12^\circ = \boxed{78^\circ}. ]
