1. ## Diagonals in a quadrilateral

In a converx quadrilateral $ABCD$, points M and N are midpoints of sides AB and CD, and diagonals meet at E. Show that the line containing the bisector of angle BEC is perpendicular to line MN iff $AC=BD$.

My repeated attempts at this problem prove vain, your help would be greatly appreciated. I attach a picture which has, however, not helped me in any way.

2. Here's a vector proof that seems to work.

Denoting the bisector of BEC as EL, the scalar product of $\underline{EL}$ with unit vectors in the directions of EC and EB will be equal, so,

$\displaystyle\frac{\underline{EC}}{\mid \underline{EC}\mid}.\underline{EL}=\frac{\underlin e{EB}}{\mid\underline{EB}\mid}.\underline{EL}\ \ \dots\dots(1)$

Now,

$\displaystyle\underline{MN}=\underline{MA}+\underl ine{AC}+\underline{CN}$

and

$\displaystyle\underline{MN}=\underline{MB}+\underl ine{BD}+\underline{DN}$,

so adding, and dividing by 2,

$\underline{MN}=\frac{1}{2}(\underline{AC}+\underli ne{BD})=\frac{1}{2}(\lambda\underline{EC}-\mu\underline{EB}),$ for some constants $\lambda$ and $\mu.$

If MN and EL are to be at right angles to each other, their scalar product will have to equal zero so, using (1),

$\lambda$ will have to be some multiple of $1/\mid\underline{EC}\mid$ and $\mu$ will have to be the same multiple of $1/\mid\underline{EB}\mid$ implying in turn that $\underline{AC}$ and $\underline{BD}$ are equal in magnitude.

3. Hello atreyyu
Originally Posted by atreyyu
In a converx quadrilateral $ABCD$, points M and N are midpoints of sides AB and CD, and diagonals meet at E. Show that the line containing the bisector of angle BEC is perpendicular to line MN iff $AC=BD$.

My repeated attempts at this problem prove vain, your help would be greatly appreciated. I attach a picture which has, however, not helped me in any way.

On your diagram, construct the bisector of $\angle AEB$, meeting AB and CD at P and Q, respectively. Then, since this bisector is perpendicular to the bisector of $\angle BEC$, we need to prove that $PQ \parallel MN$ iff AC = BD.

Let the points of intersection of MN with AC and BD be R and S respectively, and suppose that:
$AM = MB = m$

$DN=NC=n$

$AE = b$

$BE = a$

$EC=d$

$ED =c$
Then, using the angle bisector theorem on $\triangle AEB$
$\dfrac{AP}{AE}=\dfrac{PB}{EB}$

$\Rightarrow AP = \dfrac{b}{a}PB$

$\Rightarrow AP +PB = 2m = \left(\dfrac{a+b}{b}\right)PB$

$\Rightarrow PB = \dfrac{2ma}{a+b}$

and $AP = \dfrac{2mb}{a+b}$

Similarly
$DQ = \dfrac{2nc}{c+d}$

$QC= \dfrac{2nd}{c+d}$

Now we'll assume that $PQ \parallel MN$ and prove that $AC =BD$. I'll leave you to prove it the other way around.

So, from $\triangle DEQ$, $NS\parallel QE$:
$\dfrac{DN}{DQ}=\dfrac{DS}{DE}$

$\Rightarrow \dfrac{n(c+d)}{2nc}=\dfrac{DS}{c}$

$\Rightarrow DS = \frac12(c+d)$

Again, in a similar way, from $\triangle BMS$:
$BS = \frac12(a+b)$

But $BS + SD = BD = a+c$
$\Rightarrow \frac12(a+b) +\frac12(c+d)=a+c$

$\Rightarrow a+ b+c+ d = 2a+2c$

$\Rightarrow b+d = a+c$

$\Rightarrow AC = BD$