BdMO National Higher Secondary 2008/9

[Moon]

Problem 9:
Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$. Note, $AC \not = BD$. Let $E$ be the intersection point of the diagonals of $ABCD$. $AE = DE$. If $\angle BAD+\angle ADC=\theta$ , find $\theta$.

[bristy1588]

Hints anyone?

[Fatin Farhan]

Someone..?

[*Mahi*]

Hints

The answer to this question is $120^\circ$

Let $AB= a$. Let the circle $(A, a)$ meet $DE$ at $B, B'$ and let the circle $(D, a)$ meet $AE$ at $C, C'$.
Let wlog $BD>AC$.
$B, B', C, C'$ concyclic on a circle with radius $a$. Let the center be $A'$
$A',A$ symmetric wrt $BB'$, $A'D$ symmetric wrt $CC'$
Then prove $\angle A'AD = \angle A'DA = 60^\circ$
Which implies $\angle EAD = \angle EDA = 30^\circ$. Use this to find out the final answer.

[asif e elahi]

Let $\angle BAC=x$,$\angle CBD=y$.
$\frac{BE}{DE}=\frac{BC\times sin \angle BCE}{DC\times sin \angle DCE}=\frac{sin x}{sin (180^{\circ}-2y-x)}=\frac{sin x}{sin (2y+x)}$
or $\frac{BE}{AE}=\frac{sin x}{sin (2y+x)}$
or $\frac{sin x}{sin (180^{\circ}-2x-y)}=\frac{sin x}{sin 2y+x}$
or $\frac{sin x}{sin (2x+y)}=\frac{sin x}{sin (2y+x)}$
or $sin (2x+y)=sin (2y+x)$
So $2x+y=2y+x or 2x+y+2y+x=180^{\circ}$
The 1st case implies $x=y$.Then $ABCD$ will be a cyclic quad and $AC=BD$
So $2x+y+2y+x=180^{\circ}$
or $x+y=60^{\circ}$
then $\theta =\angle BAD+\angle ADC=\angle BAC+\angle CDB+\angle EAD+\angle EDA$
$=x+y+\angle BEA=60^{\circ}+\angle EBC+\angle ECB=60^{\circ}+x+y=120^{\circ}$