Let $ABC$ be a triangle with incentre $I$. A point $P$ in the interior of the triangle satisfies $\angle PBA+\angle PCA=\angle PBC+\angle PCB$. Show that $AP\geq AI$, and the equality holds if and only if $P=I$.
Please help me. I couldn't get anything from the question..
I will only give a brief explanation to the solution of this problem.
Referring to the diagram below,
we need the following knowledge:- Let I be the in-center of $\triangle ABC$. The perpendicular bisector of BC and the angle bisector of $\angle A$ will meet at X and X is right on circumference of the circle ABC. In addition, X happens to be the center of the circle passing through B, I, C. Then, XB = XI = XC.
Based on the given, after some calculation, we get:-
∠IBP = ∠IBC - ∠PBC = 1/2 ∠ABC - ∠PBC = 1/2 [∠PBA - ∠PBC] …. (1)
Similarly, ∠ICP = ∠PCB - ∠ICB = ∠PCB - 1/2 ∠ACB = 1/2[∠PCB - ∠PCA] …. (2)
Since ∠PBA +∠PCA = ∠PBC +∠PCB , then ∠PBA - ∠PBC = ∠PCB - ∠PCA …. (3)
(1) , (2) and (3) imply: ∠IBP = ∠ICP and therefore BIPC is cyclic. This is equivalent to adding P as another con-cyclic point to the circle BIC. See the diagram below,
In $\triangle PAX$, by triangle inequality, we have $AP + PX \ge AX = AI + IX$
Result follows after subtracting $PX = IX$ from both sides.
The solution given at the linked website is not clear and also wrong, so I'll write a clearer and correct solution here. To really understand this, you probably need to draw a figure (as I did when writing this).
Let $J$ be the circumcenter of $\triangle CIB.$ The key part of this problem is proving that $IJ$ is parallel to $AI.$ Since $JC,$ $JB,$ and $JI$ are radii of the circumcircle, their lengths are all equal to the same value (that is, $|JC|=|JB|=|JI|=R_{CBI}$). Triangles $JCI,$ $JBI,$ and $JCB$ are isoceles, so $\angle JCI=\angle JIC,$ $\angle JBI = \angle JIB,$ and $\angle JBC=\angle JCB.$ Using the fact that $I$ is the intersection of the angle bisectors of $\triangle ABC,$ the sum of the interior angles of quadrilateral $CIBJ$ is $$ \begin{align} & 2(C/2+\angle JCB) + 2(B/2+\angle JCB) + (\pi - 2\angle JCB) \\ & \qquad = C+B + 2\angle JCB + \pi \\ & \qquad = \pi-B-A + B + 2\angle JCB + \pi \\ & \qquad = 2\pi - A + 2\angle JCB \\ & \qquad = 2\pi \end{align} $$ so $\angle JCB = A/2$ and $\angle JIB=A/2+B/2,$ which is $\pi - \angle AIB.$ So $AI$ is parallel to $IJ.$
Let $P=I.$ Then $\angle PBA=B/2,$ $\angle PCA = C/2,$ $\angle PBC=B/2,$ and $\angle PCB=C/2$ (so the condition of the problem is satisfied). When $P=I,$ $\angle PBA=\angle PBC$ and $\angle PCA=\angle PCB.$ Moving $P$ from $I$ means setting $\angle PBA = B/2-s,$ $\angle PBC=B/2+s,$ $\angle PCA=C/2-t,$ and $\angle PCB=C/2+t$ for some angle increments $s$ and $t.$ The condition $\angle PBA + \angle PCA = \angle PBC + \angle PCB$ is now equivalent to $s=-t.$ So when decreasing $\angle PBA$ by $s$ (for some $s\in (-B/2,B/2)$), $\angle PCA$ must be increased by $s.$ This means $\angle PBI = \angle PCI = s.$ So a classical geometry theorem says that quadrilateral $BPIC$ is cyclic, meaning $P$ is on the circumcircle of $\triangle CBI,$ meaning $|JP| = |JI|.$
Finally, the triangle inequality applied to $\triangle APJ$ shows that $|AP|+|PJ| \geq |AJ|=|AI|+|IJ|.$ Since $|PJ| = |IJ|,$ $|AP|\geq |AI|$ (with equality only when $\triangle APJ$ is degenerate, that is, when $P=I$).