Simple circle geometry/ similarity question How would you prove that $a=b$ ?
Would it be possible to solve this using similarity or trigonometry?

Thank you in advance for any help. Any theorems or links would be appreciated.
 A: 
$\angle ABD+\angle AGD=\pi$ $\Longrightarrow$ $\angle ABC=\angle AGD$ $\Longrightarrow$ $\angle AEC=\angle AFD~$ and $~\dfrac{AE}{EC}=\dfrac{AF}{FD}$
$\therefore$ $\triangle AEC\sim\triangle AFD$
A: I take the picture to illustrate this scenario: Circles $\bigcirc X$ and $\bigcirc Y$, of respective radii $r$ and $s$, intersect at $P$ and $Q$. A line through $Q$ meets the circles at $A$ and $B$.

Now, $\triangle XAP$ and $\triangle YBP$ are isosceles, with
$$\frac{|XA|}{|YB|} = \frac{|XP|}{|YP|} = \frac{r}{s}$$ 
If we can show also that 
$$\frac{|AP|}{|BP|} = \frac{r}{s} \qquad (\star)$$
then the triangles will be similar, and their base angles, congruent.
Recall that the Law of Sines relates angles and sides of a triangle, as well as the diameter of the circumcircle. So, for instance, in $\triangle APQ$,
$$\frac{|PQ|}{\sin\angle PAQ} = \frac{|AP|}{\sin\angle AQP} = \frac{|AQ|}{\sin\angle APQ} = 2 r$$
Of importance to us is that
$$|PQ| = 2r \sin\angle PAQ$$
Likewise, in $\triangle BPQ$,
$$|PQ| = 2s \sin\angle PBQ$$
We now apply the Law of Sines to $\triangle APB$ to get
$$\frac{|AP|}{|BP|} = \frac{\sin\angle PBQ}{\sin\angle PAQ} = \frac{|PQ|/(2s)}{|PQ|/(2r)} = \frac{r}{s}$$
proving $(\star)$.
A: *

*LET THE CENTRE OF THE TWO TRIANGLE ARE  O & P

*now a  =$\frac{180-t}{2}$;  now t=2l

*SO a=$\frac{180-2l}{2}$

*NOW again b =$\frac{180-x}{2}$; x=2f and f=180 $-$n 

*SO b =$\frac{2n-180}{2}$

*we see (n+l)=180; FROM THIS WE GET $\frac{180-2l}{2}$ =$\frac{2n-180}{2}$

*so a=b
