How to find the length of the line segment of this following triangle?

Question

Given: right triangle ABC with C as a right angle. Line DE is formed, such that ED is perpendicular to AB. D at AB and E at CB.

If AC=AD= 8 cm and CE=DE, what is the length of BE?

My thinking(since I didn't have any idea to solve this problem) :

I think we need to find out the length of all the triangle legs to find out the length of BD and BC.

From the fact that CE=DE, I can guess that the shape of ACED is kinda like a kite, so ACE is congruent to ADE.

Since, Triangle ACE is congruent to triangle ADE, so the angle CAE is congruent to angle EAD too.

But, this observation cannot help me to find out the certain length.

Can somebody provide more hint for me to solve this problem?

Thanks

Answer 1

Because $\Delta ACE$ and $\Delta BDE$ are similar, we have:

$$\dfrac{|BE|+|CE|}{|AC|}=\dfrac{|BD|}{|DE|}$$

We also have:

$$\dfrac{|AD|+|BD|}{|AC|}=\dfrac{|BE|}{|DE|}$$

and so:

$$\dfrac{|BE|+|CE|}{|BD|}=\dfrac{|AD|+|BD|}{|BE|}$$

Using $|AC|=|AD|=8$ and $|CE|=|DE|$ this becomes:

$$|BE|(|BE|+|CE|)=|BD|(8+|BD|)$$

We also have that $|BD|^2+|CE|^2=|BE|^2$, so we arrive at a final equation in $|BE|,|CE|$. But $|CE|$ is (reasonably) any value, so there is no fixed value to $|BE|$.

Answer 2

I do not think that the length of BE is determined completely by the data. For one thing, the fact that CE=DE already follows from the other data. On the other hand, one can freely vary the magnitude of the angle CAE=EAD while keeping AC=AD constant at 8 cm, and still obtain a variety of lengths for BE. Try a hand drawing of the two extreme cases where CAE=EAD is very sharp (2°) and where CAE=EAD is nearly a half straight angle (43°)