Problem: Let $D$, $E$ and $F$ be the points of tangency between the incircle of triangle $ABC$ and the sides $BC, CA,$ and $AB$. Let $M$ and $N$ be the feet of the perpendiculars from $B$ onto $CI$, and from $C$ onto $BI$, respectively. Prove that the points $M, D, E,$ and $N$ are collinear.

So Far I have:

Since $BMDI$ is a cyclic quadrilateral this implies that $\angle BMI = 90^\circ = \angle BDI$.

We know that $\angle MDI + \angle MBI = 180^\circ$. Denote $\angle MDB = \theta$. Then $\angle MBI = 90^\circ - \theta$.

If we can only prove: $\angle MDI + \angle IDE = 180^\circ$, then we can prove $M, D, E$ are collinear. (and then use a similar method to prove $D, E, N$ collinear).

$\triangle BDI \equiv \triangle BFI$ by HL so $\angle DBI = 90^\circ - \theta - \beta = \angle FBI$ ($\beta = \angle MBD$) and $\angle DIB = \angle FIB = \theta + \beta$.

Any help on what I should do next is appreciated![enter image description here]1


It is enough to prove that $M,D,E$ are collinear. After that we can use the same idea and get that $D,E,N$ are also collinear and then $M,D,E,N$ are collinear.

First of al we have that $BIDM$ is cyclic because $\angle BMI=\angle BDI (=90°)$, so $$\angle MDI=180°-\angle MBI.\quad (*)$$

We then need to prove that $\angle IDE=\angle MBI$.

$1)$ By sum of angle in $\Delta BIC$ we get $\angle BIC=180°-\frac{\angle ABC+ \angle ACB}{2}=90°+\frac{\angle BAC}{2}$. But we also have that, by exterior angle theorem at $\Delta BMI$, $\angle BIC=90°+\angle MBI$, so $$\angle MBI=\frac{\angle BAC}{2}$$.

$2)$ In the quadrilateral $ADIE$ we get $\angle DIE=180°-\angle BAC$ and from the isosceles triangle $IDE$ we get $$\angle IDE=\frac{1}{2}(180°-\angle DIE)=\frac{\angle BAC}{2}$$.

And then, from $(1)$ and $(2)$ we get $\angle IDE=\angle MBI$ and backing to $(*)$ we get:

$$\angle MDI+\angle IDE=180°-\angle MBI+\angle MBI=180°$$

so, $M,D,E$ are collinear.

  • $\begingroup$ How did you determine that $\angle BIC - 180^\circ - \displaystyle\frac{\angle ABC + \angle ACB}{2}$ @Arnaldo $\endgroup$
    – rover2
    May 17 '17 at 16:48
  • $\begingroup$ @rover2: sum of angles in $\Delta BIC$ $\endgroup$
    – Arnaldo
    May 17 '17 at 16:51
  • $\begingroup$ so due to the fact that $\triangle BDI \equiv \triangle BFI$ implying that $\angle DBI = \angle FBI$ that means that $\angle FBI = \frac{1}{2} \angle ABC$ right? $\endgroup$
    – rover2
    May 17 '17 at 16:59
  • $\begingroup$ @ you are right! $\endgroup$
    – Arnaldo
    May 17 '17 at 17:03
  • 1
    $\begingroup$ @Sorfosh: Ok, sorry. It is because $\angle ADI +\angle AEI=90°+90°=180°$ $\endgroup$
    – Arnaldo
    May 17 '17 at 22:55

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