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Regards SD ----------------------------- Press Kudos if you like my post. Debrief 610-540-580-710(Long Journey): http://gmatclub.com/forum/from-600-540-580-710-finally-achieved-in-4th-attempt-142456.html
I thought this was a very difficult problem. I stared at this for a while before figuring it out. Make point X the point where DB and AF intersect. The angles of the four sided figure in the middle (BCFX), add up to 360. If there are two 90 degree angles, then the remaining angles, FXB and FCB, add up to 180. You will also note that angles FXB and FXD also add up to 180 as well as angles FXB and AXB.
FXB + FCB = 180 FXB + FXD = 180 FXB + AXB = 180
Therefore, Angles FCB, FXD, and AXB are all equal. Lets call their value Y for now. Now you have 3 triangles-- AXB, DXF, AFC, and DBC-- that all have angles that are 90 and Y degrees. They all have the same 3rd angle, which we can call Z for now. We do not need to know the value of Y or Z.
Now, from the above configuration lets extract 2 triangles: CFA and CBD. We can find their proportionality be comparing known values on proportional sides. Since we have values for both CA and CD and they proportional, lets use them. CB:CD = (12):(18) = 2:3. Therefore, the ratio of the sides of triangle CFA to CBD is 2:3. Now lets find AF. Since AF and BD are proportional at 2:3 and BD = 15, then AF = 15(2/3) = 30/3 = 10
Such figures should immediately remind you of similar triangles. The two triangles FCA and BCD have a right angle each and a common angle (angle A) each. Hence, their third corresponding angles would also be equal to each other. By AA rule, the triangles are similar. Therefore, the ratio of their corresponding sides will be equal.