I need urgent help with this geometry question ASAP.?

Question

There is a triangle ABC with base BC. The side BA is extended to D and the side BC is extended to E such that AE bisects angleCAD. Find the length of CE, is AB = 10cm, BC = 12cm and AC = 6cm.

I know it has something to do with similarity of triangles but I don't know the method to get the answer. Please help ASAP.

Answer 1

Draw the picture, and make another line from E to D. Since angle EAC = angle CAB, AC is parallel to DE and the triangles are similar, which means their sides will be in proportion. Let me work on this a little more before the man gets hold of it....
ok, I'm back...
If you look at the picture some more, you'll see that, because of parallel lines, angle AED = angle EAD = angle EAC so triangle ADE is iscocolese. Let x = the length that BA was extended to D. Then by similar triangles, 10/(10 + x) = 6/x. Cross multiply and 10x = 6(10 + x). 10x = 60 + 6x, 4x = 60, x = 15. Remember, x is the amount you've extended AB by so you still need to find CE. My brother needs the computer so I'll be back or someone else will finish....

Answer 2

There is nothing in this problem that pinpoints the location of D. So any line segment involving D should not be part of the solution. The answer seems to be that CE = 18 cm, but I'm still trying to figure out how to prove it.

Answer 3

Thales' Theorem:
AB/BC = CE/AC
6/12 = CE/10
CE = (6 X 10) : 12
CE = 5 cm
Please check it.
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Answer 4

I got about 17.7.
I used the law of sines and cosines.
You can find angle BAC and ACB using law of cosines.
Then angle CAE is (180-BAC)/2.

Then use the law of sine for rest of problem.