Problem E : 6411 by Mihai MICULIȚA

            Mihai MICULIȚA was a prolific author of geometry problems. In the 1980s (when he was active in Oradea) he focused more on geometry problems in space. 

          You can find magazines like the one below here and here.

On Page 553, two of my College colleagues appear among the solvers, BAKONYI Mihai from ARAD and MATEI NEDA Elena (later married GHILERDEA) from BOZOVICI.

On page 521 :

"E : 6411.  Let  $ABC$ be a triangle in which  $AB\neq AC$. If  $D$  is the foot

 of the altitude drawn from  $A$ , and  $M, \;N\; and\; P$  the midpoints of the sides

  $AB,\; AC\; and\; BC$  respectively, and  $O$ is the midpoint of the segment  $DP$ ,

            prove that the triangle  $MON$  is isosceles.

 {Author : } Mihai MICULIȚA, teacher, Șimleul Silvaniei"

We will apply what was discussed here regarding Geometric Objects to solve the problem.

ANSWER CiP

$OM=ON$

                                   Solution CiP

         Let's complete the figure with a few additional lines.

          In this way we will identify several important Geometric Objects: $O_1$ - The Midline in the triangle and   $O_2$  The Median on the Hypotenuse in the right triangle.

         Object O1 

 has the properties : 

-  $AM=MB=\frac{AB}{2}\;\;\;,\;\;AN=NC=\frac{AC}{2}$

-  $MN \parallel BC$

-  $MN=\frac{BC}{2}=BP=PC$

-  $\widehat{ABC}=\widehat{AMN}\;\;\;,\;\;\widehat{ACB}=\widehat{ANM}$

This object appears three times in the above Figure of the problem : $MN\parallel BC\;\;,\;\;NP\parallel AB\;\;,\;\;MP\parallel AC$

Object O2

has the  properties : 

-  $\widehat{ADC}=90^{\circ}\;\;,\;\widehat{DAC}+\widehat{ACD}=90^{\circ}$

-  $AM=\frac{AC}{2}=MC=DM$

-  $\widehat{ADM}=\hat{A}\;\;,\;\;\widehat{CDM}=\hat{C}$

-  $\widehat{AMD}=2\cdot \hat{C}\;\;,\;\;\widehat{CMD}=2\cdot \hat{A}$

         Another object appears, which may go unnoticed : it is the 

isosceles trapezoid

with the  properties :  

-  $DP\parallel MN$

-  $PN\not\parallel DM\;\;,\;\;PN=DM$

-  $\hat{N}=\hat{M}\;\;,\;\;\hat{P}=\hat{D}$

-  $\hat{D}+\hat{M}=180^{\circ}\;\;,\;\;\hat{P}+\hat{N}=180^{\circ}$

-  $PM=DN$

          In the figure completed from the solution we have :

$$NP\underset{O_1}{=}\frac{AB}{2}\underset{O_2}{=}DM$$

so the isosceles trapezoid MNPD immediately appears. I was very proud of this configuration, I used it at least once at the Teacher Certification Competitions.

     Then immediately the key to solving the problem appears, because starting from the last object we have the next one

It is immediate that, with the SAS case, we have the equality of triangles

  $\Delta NPO=\Delta MDO$

  from which we find  $OM=ON$

$\blacksquare$