Prove that $MN$ and $ED$ are parallel

Question

Let $ABC$ be a triangle with the measure of angle $ABC=45$, $AB> AC.$ If $H$ is the orthocenter of triangle, $M,N$ are midpoints of $(AH), (BC)$ and $E,D$ are projections of $B$ and $A$ respectively on the opposite sides of triangle, prove that $MN$ and $ED$ are parallel.

I considered $F$ midpoint of $AB$ and tried using parallelism for searching some congruent angles. The triangle $HDC$ is isosceles and the angle $EDC$ is congruent with angle $BAC$, so I want to prove that angle $MND$ is congruent with $BAC$.

Answer 1

Let's place your figure on the Cartesian coordinate system. Based on the fact that $\angle ABC= 45^{\circ}$, we may assume that:

$$B=(-1,0) \\ A=(0,1) \\ D=(0,0),$$

and $C=(a,0).$ So, the equation of $AC$ is $y=\frac{-1}{a} x+1$, and consequently that of $BE$ is $y=ax+a.$ Therefore, regarding the coordinates of $H$ (as the intersection of $BE$ and $AD$) and $E$ (as the intersection of $BE$ and $AC$), we get:

$$ E=(\frac{a-a^2}{1+a^2},\frac{a+a^2}{1+a^2}),$$ and $$ H=(0,a).$$

As a result,

$$M=(0, \frac{1+a}{2}).$$

On the other hand, $N=(\frac{a-1}{2},0).$

Now, to be done, we only need to show that:

$$\frac{\frac{1+a}{2}-0}{0-\frac{a-1}{2}}= \frac{\frac{a+a^2}{1+a^2}-0}{\frac{a-a^2}{1+a^2}-0},$$

which is correct.

Answer 2

Hint: You can use the properties of nine point circle( call it $\Omega$).The center G of $\Omega$ is the mid point of OH, where O is the center of circumscribe circle of ABC. M is Euler point and MN is the diameter of $\Omega$.Then show that $FM=FN$, this says that a perpendicular from F on MN is another diameter of $\Omega$ which crosses G ,the center of $\Omega$. You showed $\angle EDC=\angle BAC$. Triangles DEC and ABC have a common angle and one equal angle, so their third angles are equal that is $\angle DEC=\angle ABC=45^o$.Connect G to E and D. Show $GD||AC$ and get result that $\angle HDE=45^o$. Triangle GDE is isosceles so $\angle GCD=45^o$which means $FG\bot DE$, that means $FG\bot MN$ and $MN||DE$.

Answer 3

Let us recall some known facts related to the nine-point-circle (or Euler circle) of a general triangle $\Delta ABC$.

Let $O,H,9$ be the circumcenter, the orthocenter, and the mid point of $OH$, the center of the nine point circle, the Euler circle passing through the points $A',A'', A_1; B',B'',B_1;C',C'',C_1$. Here $A'$ be the mid point of the side opposite to $A$, $A''$ the projection of $A$ on this side, and $A_1$ the mid point of $AH$. The other points are constructed similarly. A first property is that $A_1HA'O$ is a parallelogram. (Its diagonals intersect in their mid points.) So $AA_1=A_1H=OA'$, showing that $AA_1A'O$ is also a parallelogram. (Sides $AA_1$ and $OA'$ are parallel and congruent.) So we record a main direction of lines used in the sequel: $$ A_1A'\|AO\ . $$ Let us construct as in the figure the point $P$ as the intersection of $C'O$, (the perpendicular bisector of $AB$,) with the line through $B''$ having the above direction. (So the direction is $B''P\|A_1A'\|AO$.) Then: $$ \widehat{C'PB''} = \widehat{C'OA} = \frac 12\widehat{BOA} = \widehat{BCA} = \widehat{B''C''A} \ , $$ showing $B''C''C'P$ cyclic. Three vertices are on the nine-point-circle $(9)$, so $P$ is also on this circle. Note that $P$ is the opposite of $C''$ on this circle, since the direction $PB''\|OA\|A'A_1$ is perpendicular on $B''C''$. (To see this look at $\widehat{AOC'}=\widehat{B''C''A}$, and isolate the resulted cyclic quadrilateral with a right angle in $C'$, so its opposite angle in it is also a right angle.)

Construct $Q$ similarly on the circle $(9)$, which turns out to be opposite to $B''$ on this circle.

Question: When do we have $P=A''$?

(This is the posted question in one direction.)

This is equivalent to having $B''A''\|A_1A'$. And equivalent to having $A''B''$ perpendicular on $B''C''$, i.e. $$ \widehat{A''B''C''} \overset?=90^\circ\ . $$ Since the angle in $B$ induces two congruent angles in $B''$, the violet ones in the picture, we conclude, in part recalling the above: $$ PB''\|A'A_1\Longleftrightarrow A''=P\Longleftrightarrow A''B''\perp C''B''\Longleftrightarrow \hat B=45^\circ\ . $$ $\square$

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In the above picture, the angle in $C$ is rather closer to $45^\circ$, making $Q$ relatively close to $A''$.