# Using Menelaus to show: The perpendicular bisectors of the angles bisectors of a triangle meet opposite sides in collinear points

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.

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.

• 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.
– Blue
Jun 7, 2018 at 18:09

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

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. $\square$