Tough Puzzle: Ratio of similar triangles

[The_Big_K]

Puzzle Source: Mind and Visual Puzzles!

If the segment A'B' is tangent to the inscribed circle of triangle ABC, and that segment AB = segment CM; then, what is the ratio of the area of the triangle ABC to the area of the small triangle A'B’C? Hint: angle CAB is not necessarily a right angle, however triangles ABC and A'B'C are similar!

[The_Big_K]

21 views so far and not a single attempt?

Too tough for you, CEans?

All the best!

[The_Big_K]

Where did all the talented people on CE go?

Come on its simple Geometry. Did you not learn it in school?

[The_Big_K]

Woo hoo.

I can't believe this! CEans can't crack a simple Geometry problem? . Where are the engineers who solve the toughest problems in the world! Where are they?

Bring them on CE! Come on! Oh Lord!

[satheesh27887]

HELLO... ''THE BIG K''
we don't know how to begin....
Can u give us another hint.....

[The_Big_K]

No, sorry. I'd have given hints if anyone had attempted it on the day it was posted.

Now there are only two options - either crack the puzzle or accept the defeat!

[KSHIRABDHI]

i think the ratio of triangle ABC to triangle A'B'C is 2:1. as both the triangles are similar so considering AB as base and A'B' is parallel to AB then as per similarity rule the ratio is 2:1

[yudi]

i think the ratio Triangle ABC/A'B'C should be eqaul to (A'C/AC)^2
taken AB as the base of triangle ABC & A'B'||AB as ABC~A'B'C

[banu]

abc:a'b'c is 2:1

[satheesh27887]

Originally Posted by The_Big_K

No, sorry. I'd have given hints if anyone had attempted it on the day it was posted.

Now there are only two options - either crack the puzzle or accept the defeat!

k.... i accept the defeat.... we r eager to know the answer.......