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 $\displaystyle BS + SD = BD = a+c$

$\displaystyle \Rightarrow \frac12(a+b) +\frac12(c+d)=a+c$

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

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

$\displaystyle \Rightarrow AC = BD$

Grandad