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Prove that the perpendicular bisectors of the interior angle bisectors of any triangle meet the sides opposite the angles being bisected in three collinear points.

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Here is what I have so far. In order to show H, G, and I are collinear, I need to show that $$\frac{AH}{HB}\cdot\frac{BI}{BC}\cdot\frac{CG}{GA}=-1$$ This is true by Menelaus's Theorem.

Now, I can prove by SAS that $\Delta FJO\cong \Delta CJO$ and by ASA that $\Delta OJC\cong \Delta PJC$. Thus, by transitivity, we have $\Delta OJC \cong \Delta PJC \cong \Delta OJF$. Therefore, $\angle PCJ\cong\angle OFJ$. This implies that $OF\parallel BC$ since the alternate interior angles are congruent. I know that parallel lines divide the sides of the triangle proportionally, so we get $\frac{AF}{FB}=\frac{AO}{OC}$.

Through similar procedures, I can also conclude the following: $FP \parallel AC$ which implies $\frac{BP}{PC}=\frac{BF}{FA}$ and $ME \parallel BC$ which implies $\frac{AM}{MB}=\frac{AE}{EC}$.

I also know there are three simple results from the angle bisector theorem. $$\frac{BD}{AB}=\frac{CD}{AC}$$ $$\frac{AF}{AC}=\frac{BF}{BC}$$ $$\frac{AE}{AB}=\frac{EC}{BC}$$

I also know from Ceva's Theorem that since the angle bisectors are concurrent, then $$\frac{AF}{FB}\cdot \frac{BD}{DC} \cdot \frac{CE}{EA}=1$$ Though that piece of information is not really different from the results of the angle bisector theorem.

I need to be able to combine things somehow, but I'm not seeing a connection between what I know and the points H, I, and G.

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  • $\begingroup$ Your statement of Menelaus' Theorem is incorrect. It should be $$\frac{AI}{IB}\cdot\frac{BG}{GC}\cdot\frac{CH}{HA}=-1$$ Here, it's actually easier to use the trigonometric form: $$\frac{\sin\angle ACI}{\sin\angle ICB}\cdot\frac{\sin\angle BAG}{\sin\angle GAC}\cdot\frac{\sin\angle CBH}{\sin\angle HBA}=-1$$ A little angle chasing shows that, in your diagram, $\angle ACI = \angle B$, $\angle BAG = \angle C$, $\angle ABH = \angle C$. The other angles are easy to find, and their sines "obviously" cancel appropriately. $\endgroup$
    – Blue
    Jun 7, 2018 at 18:09

1 Answer 1

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Let $\overline{AD}$ be the bisector of $\angle A$ of $\triangle ABC$, and let $\overline{MP}$ be the perpendicular bisector of $\overline{AD}$.

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Because $P$ is on perpendicular bisector, we have $\angle DAP \cong \angle PDA$ at the base of isosceles $\triangle PAD$. But $\angle PDA$ is an exterior angle of $\triangle ABD$, so that $\angle PDA = \frac12\angle A + \angle B$. Since $\angle DAC$ is also $\frac12\angle A$, we deduce that $$\angle CAP = \angle B \phantom{+\angle C}\qquad\qquad \angle PAB = \angle A + \angle B$$

Likewise, $$\begin{align} \angle ABQ &= \angle A + \angle B \qquad\qquad \angle QBC = \angle A \phantom{+\angle C} \\[4pt] \angle BCR &= \angle A \phantom{+\angle C\;} \qquad\qquad \angle RCA = \angle C + \angle A \end{align}$$

(The betweenness of $P$, $Q$, $R$ relative to the the vertices of $\triangle ABC$ can change which angles shown directly match $\triangle ABC$'s angles, and which are sums of those angles, but the above is typical.)

Invoking the trigonometric form of Menelaus' Theorem, and providing explicit "$-$"s to recognize that the angles in each ratio are oppositely oriented, we have

$$\begin{align} -\frac{\sin\angle CAP}{\sin\angle PAB} \cdot -\frac{\sin\angle ABQ}{\sin\angle QBC} \cdot -\frac{\sin\angle BCR}{\sin\angle RCA} &= -\frac{\sin B}{\sin(A+B)} \cdot \frac{\sin(A+B)}{\sin A} \cdot \frac{\sin A}{\sin(C+A)} \\[4pt] &= -\frac{\sin B}{\sin(180^\circ - B)} \\[4pt] &= -1 \end{align}$$

Therefore, $P$, $Q$, $R$ are collinear. $\square$

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