I am stumped on the following question:
In the figure below $AD=4$ , $AB=3$ , and $CD=9$. What is the area of Triangle AEC?
I need to solve this using trigonometric ratios however if trig. ratios makes this problem a lot easier to solve I would be interested in looking how that would work.
Currently I am attempting to solve this in the following way
$\bigtriangleup ACD_{Area}=\bigtriangleup ECD_{Area} + \bigtriangleup ACE_{Area}$
The only problem with this approach is finding the length of AE or ED. How would I do that ? Any suggestion or is there a better method
Best Answer
Hint: $\bigtriangleup DCE \sim \bigtriangleup ABE$ as $BC$ is crossing parallel lines.