1. ## Diagonals in a quadrilateral

In a converx quadrilateral $\displaystyle ABCD$, 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 $\displaystyle AC=BD$.

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 $\displaystyle \underline{EL}$ with unit vectors in the directions of EC and EB will be equal, so,

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

Now,

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

and

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

so adding, and dividing by 2,

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

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

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

3. Hello atreyyu
Originally Posted by atreyyu
In a converx quadrilateral $\displaystyle ABCD$, 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 $\displaystyle AC=BD$.

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

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

$\displaystyle DN=NC=n$

$\displaystyle AE = b$

$\displaystyle BE = a$

$\displaystyle EC=d$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle \Rightarrow AC = BD$