3

Let $K,M,N$ be feet of the altitudes of $\triangle ABC$ from the vertices $A,B,C$ respectively, $P$ be the midpoint of the edge $\overline{AB}$ and $R$ be the intersection point of the lines $AB$ and $KM$. Prove: $$|RA|\cdot|RB|=|RP|\cdot|RN|$$

My attempt:

Edit: In the meantime, I improved my proof a little bit.

Let $O$ be the orthocenter of $\triangle ABC$. Then $\triangle MKN$ is orthic.

Since $\measuredangle BMA=\measuredangle BKA=90^\circ$, $ABKM$ is a cyclic quadrilateral.

According to the power of a point theorem: $$|AR|\cdot|BR|=|MR|\cdot|RK|.$$

Feet $K,M,N$ of the altitudes and the midpoint $P$ are concyclic,i. e., $K,M,N, P$ belong to the nine-points circle, so we can apply the power of a point theorem once again: $$\begin{aligned}&|RN|\cdot|RP|=|MR|\cdot|RK|\\\implies& |AR|\cdot|BR|=|RN|\cdot|RP|\end{aligned}$$

However, I don't remember we have mentioned the nine-points circle in our online lectures and I don't know how to prove $K,M,N,P$ are concyclic.

I tried to prove $\triangle NMR\sim\triangle KRP$, but I'm not sure if it really is so.

Quadrilaterals $ANOM, NBKO$ and $CMOK$ are cyclic. $$\begin{aligned}|PB|=|PM|\implies\measuredangle PKM&=\measuredangle PBK\\\measuredangle OCM&=\measuredangle OKM=\measuredangle ABM\\\measuredangle KCO&=\measuredangle KMO=\measuredangle KAB\\\measuredangle APC&=2\measuredangle PBK\\\measuredangle PKA+\measuredangle BKP&=90^\circ\\\implies\measuredangle PKM&=\measuredangle PBK\end{aligned}$$

enter image description here

Just in case: $\color{red}{\triangle AKC\sim\triangle BCM},\color{purple}{\triangle BRM\sim\triangle KRA},\color{blue}{\triangle RBK\sim\triangle RAM}$ and $\color{green}{\triangle ABC\sim\triangle CMK}$.

Is there any way I could use those facts in this proof?

Thank you in advance!

Update:

Thanks to @Richrow in the comment section,

$$\color{red}{\measuredangle KNB}=\measuredangle KOB=\measuredangle AOM=\color{red}{\measuredangle ANM}$$

and then it follows, like @richrow said, $$\measuredangle MPK=2\measuredangle MBK$$ And special thanks to @user21820!

Matcha Latte
  • 4,665
  • 4
  • 16
  • 49
  • I'm confused about your use of the power of a point theorem. It should be $|RA| \times |RB| = |RM| \times |RK| = |RN| \times |RP|$. You have there $|RM| \times |MK|$ instead. – Todor Markov Jun 26 '20 at 10:20
  • @TodorMarkov, sorry, I see my missleading typo in the very beginning. Once I wrote it, I just copied it in the following part. – Matcha Latte Jun 26 '20 at 10:34
  • 2
    In order to prove that $K$, $M$, $N$ and $P$ are concylic you can just check the equality of $\angle MNK$ and $\angle MPK$ (they both are equal to $180^{\circ}-2\angle A$). – richrow Jun 26 '20 at 10:59
  • @richrow, thank you very much! At the end, I got lost in my own picture. – Matcha Latte Jun 26 '20 at 12:07

1 Answers1

3

Hint: Compare $∠MPA$ and $∠MKN$ by using the property of angle at circle centre and the cyclic quadrilaterals in the orthocentric system that you already found.

user21820
  • 60,745
  • 1
    I found this, $\color{red}{\measuredangle KNB}=\measuredangle KOB=\measuredangle AOM=\color{red}{\measuredangle ANM}$, but kind of got lost in my picture. Oh, I see! Thank you! – Matcha Latte Jun 26 '20 at 12:00