Intersecting Circles

Let $C',D'$ be on $\Gamma_1,\Gamma_2$ such that $BC',BD'$ are diameters respectively. $\angle BAC'=\angle BAD'=90^\circ$ , so $C'AD'$ is a straight line. $\angle BCA=\angle BC'A,\angle BDA=\angle BD'A,$ so $\triangle BCD\sim\triangle BC'D'$ $\frac{BC}{BD}=\frac{BC'}{BD'}=\frac{2*15}{2*3}=5$ .

Recall that in $\bigtriangleup ABC$ we have $\frac{a}{\sin A}=2R$ , where $R$ is the radius of its circumcircle. Now we have $BC=30\sin \angle BAC$ , $BD=6\sin \angle BAD=6\sin (180^\circ -\angle BAC)=6\sin \angle BAC$ . It follows that $\frac{BC}{BD}=\frac{30\sin \angle BAC}{6\sin \angle BAC}=5$ , as desired.

Circle chords from intersection point of two circles are at same position from centre of respective circles. Chords at same distance from centre are proportional in length to their radii. So; BC/BC=15/3=5

Here you are given the radii of the circles. And no other information like, how the circles intersect or which line through A etc. So a quick way to solve is make the diagram as simple as possible. Like, draw l is horizontal line etc. So to simplify the diagram, i imagined the circles are touching and the line is horizontal. In such case, BC:BD = 2r1 : 2r2 = 30:6. Now what you are left with is prove. What we have to prove is: BC/BD = r1/r2. When you know what you are going to prove, it is easier than before. :)

It will be true for every 2 circles. So making it a special type of circle in which, BC and BD are diameters of 2 circles. so BC = 30 and BD = 6. hence BC/ BD = 30/6 =5

Let $O_1$ and $O_2$ be the respective centers of $\Gamma_1$ and $\Gamma_2$ .

By the central angle theorem , $2\angle DCB = \angle AO_1B$ .

Notice that $\angle AO_1B$ is bisected by line $O_1O_2$ , therefore $\angle O_2O_1B = \angle DCB$ .

Analogously, $\angle O_1O_2B = \angle CDB$ .

Therefore triangle $DCB$ is similar to triangle $O_2O_1B$ , by the criteria AAA .

So $\displaystyle \frac {BC} {BD} = \frac {BO_1} {BO_2} = \frac {15} {3} = 5$ .