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

$X$ and $Y$ are the centers of the circles

Thank you in advance for any help. Any theorems or links would be appreciated.

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    $\begingroup$ Are you talking of angles $\;a,b\;$ or what? $\endgroup$ – DonAntonio Oct 26 '13 at 16:01
  • $\begingroup$ @DonAntonio yes, I am talking about the angles a and b $\endgroup$ – Berry Oct 26 '13 at 16:04
  • $\begingroup$ If the cord in the big circle were a tangent to the small circle on the upper intersection point between both circles, I think I've a solution. Are you sure this is not given in the problem, @Berry ? $\endgroup$ – DonAntonio Oct 26 '13 at 18:40
  • $\begingroup$ @DonAntonio no, that is what the diagram should be like $\endgroup$ – Berry Oct 26 '13 at 21:39
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enter image description here

$\angle ABD+\angle AGD=\pi$ $\Longrightarrow$ $\angle ABC=\angle AGD$ $\Longrightarrow$ $\angle AEC=\angle AFD~$ and $~\dfrac{AE}{EC}=\dfrac{AF}{FD}$
$\therefore$ $\triangle AEC\sim\triangle AFD$

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  • $\begingroup$ I have two comments about your solution. (1) By "exterior angle of cyclic quadrilateral", $\angle ABC=\angle AGD$ and therefore no implication like "$\angle ABD+\angle AGD=\pi$" is needed. $\endgroup$ – Mick Oct 27 '13 at 16:19
  • $\begingroup$ (2) To certain degree, your drawing seems to be requiring E, A and G to be on the same straight line. Such requirement is clearly not needed. $\endgroup$ – Mick Oct 27 '13 at 16:24
  • $\begingroup$ "by exterior angle of cyclic quadrilateral" is not proof. $\endgroup$ – chloe_shi Oct 28 '13 at 4:32
  • $\begingroup$ $G$ is arbitrary point on circumcircle of $\triangle ABD$. $\endgroup$ – chloe_shi Oct 28 '13 at 4:34
  • $\begingroup$ $G$ is for $\angle AFD$. $\endgroup$ – chloe_shi Oct 28 '13 at 4:34
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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 $B$.

enter image description here

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 $(\star)$.

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  • 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=benter image description here
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