In a Triangle, a Median is equal to an Altitude : $m_B=h_A$ // In einem Dreieck hat die Mediane die gleiche Länge wie die Höhe

I didn't discover that an altitude and a median in a triangle can be equal.

However, if  $h_A$  is equal to  $m_A$  it also appears in the Problem below. Until then, let's digress a bit.

          Let's take their formulas. Altitude :  $h_A=\frac{2}{a}\cdot \sqrt{s(s-a)(s-b)(s-c)}$

where  $s=\frac{a+b+c}{2}$ (the notations are like here).

  Median :  $m_B=\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$.

We prefer to write  $h_A=\frac{2}{a}\cdot \sqrt{\frac{a+b+c}{2}\cdot \frac{-a+b+c}{2}\cdot \frac{a-b+c}{2}\cdot \frac{a+b-c}{2}}$  or developed

$h_A=\frac{1}{2a}\cdot \sqrt{2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4} \tag{1}$

          The equality  $h_A^2=m_A^2$  is written : 

$\frac{1}{4a^2}(2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4)=\frac{1}{4}(2a^2+2c^2-b^2)\Leftrightarrow$

$\Leftrightarrow \frac{b^2}{2}+\frac{b^2c^2}{2a^2}+\not{\frac{c^2}{2}}-\frac{a^4}{4}-\frac{b^4}{4a^2}-\frac{c^4}{4a^2}=\frac{a^2}{2}+\not{\frac{c^2}{2}}-\frac{b^2}{4} \Leftrightarrow$

$\Leftrightarrow \frac{3b^2}{4}-\frac{b^4+c^4-2b^2c^2}{4a^2}=\frac{3a^2}{4}$  so in the end we get

$3a^2(b^2-a^2)=(b^2-c^2)^2 \tag{2}$

I mention that we will not use this relationship today.

          I recently saw the Problem  E:17209 in a shop window :

In translation :

               "E:17209.  In triangle ABC, median  $BM$  and altitude  $AD$  are

             concurrent at point  $P$   and have the same length. Show that  $BP = 2PD$.

 Dragoș Ionuț MATEI, Constanța"

I am attaching a figure about this problem.

There are several cases of the figure, depending on the position of point D on line BC. In any of the cases we will prove, which immediately implies the conclusion of the problem :

ANSWER CiP

$\widehat{MBC}=30^{\circ}$

                    Solution CiP (Attention, the problem is proposed for Grade 6)

              The position of point D on line BC allows us to distinguish the cases :

$(i)\;\;D=B\;;\;\;(ii)\;\;D=C\;;\;\;(iii)\;\;B-D-C\;;\;\;(iv)\;\;B-C-D\;;\;\;(v)\;\;D-B-C$

          (i)  In this case triangle  $ABC$ has a right angle at vertex  $B$.

The hypothesis  $BM=AD$  shows that  $ABM$  is an equilateral triangle, so  $CAB$ is a 30 60 90  triangle. Hence  $\widehat{MBC}=30^{\circ}$. In the equation  $BP=2PD$  all segments are zero.

          (ii)  In this case  $ABC$  has a rightangle at vertex  $C$.

From  $AD=BM=2b$(say) it follow  $MC=BM/2$  so  $BMC$ is now a 30-60-90 triangle. Because $D=C,\;P=M$  then  $BM=2MC\Rightarrow BP=2PD$.

          (iii)  The figure looks like this

We will proof this case, but we will see that the proof also applies to the others.

          We draw, both in size and orientation  $DN\overset{=}{\parallel}MC$. Quadrilateral  $CMND$  is a parallelogram.

So is  $ADMN$; in this we have

$MN=AD=BM \tag{1}$

and since  $AD\perp BC$, we also have  $MN\perp CD$. But then  $CMDN$  is a rhombus, so   $CN=ND=CM$. Here too, the line  $DC$ is the perpendicular bisector of the segment  $[MN]$, and then

$BM=BN \tag{2}$

From (1) and (2) it follows that  $BM=MN=BN$, therefore the triangle  $BMN$  is equilateral. In it, $BC$  is the bisector of the  $\measuredangle{MBN}$, therefore  $\widehat{MBC}=30^{\circ}$.

     Triangle  $BPD$  is right-angled with  $\widehat{PBD}=30^{\circ}$, so  $BP=2PD$.

          (iv)  The figure looks like this

Same proof

          (iv)  The figure looks like this

Same proof

$\blacksquare\;\blacksquare$