1. ## Help with solving.

What are the steps to solving this problem???
The second problem was written wrong. It should have been, If CD= 15, what is the length of AE?
I now understand how to do the first problem. But how do I apply it to solving the second one?
I changed the image to shows the correct question.

2. Originally Posted by Ash
What are the steps to solving these problems???
Take the problem on the left.

Triangles CBA and CED are similar since angle C is shared, angles CED and CBA are right, and angles EDC and BAC are equal.

So $\displaystyle \frac{BA}{ED} = \frac{BC}{BE}$

$\displaystyle \frac{30}{10} = \frac{CE + 30}{CE}$

$\displaystyle 3 CE = CE + 30$

$\displaystyle 2 CE = 30$

$\displaystyle CE = 15$

The second problem is supposed to be done using the same principle, but the diagram seems to conflict with the problem? (CD is the whole base, but 10 seems to be only a part of it?)

-Dan

3. Originally Posted by topsquark
Take the problem on the left.

Triangles CBA and CED are similar since angle C is shared, angles CED and CBA are right, and angles EDC and BAC are equal.

So $\displaystyle \frac{BA}{ED} = \frac{BC}{BE}$

$\displaystyle \frac{30}{10} = \frac{CE + 30}{CE}$

$\displaystyle 3 CE = CE + 30$

$\displaystyle 2 CE = 30$

$\displaystyle CE = 15$

The second problem is supposed to be done using the same principle, but the diagram seems to conflict with the problem? (CD is the whole base, but 10 seems to be only a part of it?)

-Dan
Indeed, as with the first I think he meant BA = 30, not BE.

In the second, if DE = 10, then CD can't possibly be 10, so something is wrong.

4. Originally Posted by AfterShock
Indeed, as with the first I think he meant BA = 30, not BE.

In the second, if DE = 10, then CD can't possibly be 10, so something is wrong.
On the second figure It was written wrong. It's suppose to be If CD=15 what is the length of AE?