In quadrilateral \(BCED\), sides \(\overline{BD}\) and \(\overline{CE}\) are extended past B and C, respectively, to meet at point A.

If:

**BD = 18**

**BC = 8**

**CE = 2**

**AC = 7**

and **AB = 3**...

Then what is DE?

Thanks guys

readyformathlearning Jul 26, 2024

#1**-1 **

To determine the length of \(DE\) in quadrilateral \(BCED\), we can use the properties of similar triangles and the concept of similar triangles formed by intersecting lines.

Given:

- \(BD = 18\)

- \(BC = 8\)

- \(CE = 2\)

- \(AC = 7\)

- \(AB = 3\)

We need to find \(DE\).

### Step-by-Step Solution:

1. **Identify triangles**:

- Triangles \(ABD\) and \(ACE\) share a common vertex \(A\) and extend through points \(D\) and \(E\) on \(BD\) and \(CE\), respectively.

2. **Use the intercept theorem (also known as Thales' theorem)**:

According to the intercept theorem, if two lines are intersected by a pair of parallel lines, the segments intercepted on one line are proportional to the corresponding segments intercepted on the other line.

3. **Similar triangles**:

Triangles \(ABD\) and \(ACE\) are similar by the AA criterion (angle-angle similarity) because they share angle \(A\) and both have right angles.

4. **Set up the proportionality**:

\[

\frac{BD}{DE} = \frac{AB}{AC}

\]

5. **Plug in the known values**:

\[

\frac{18}{DE} = \frac{3}{7}

\]

6. **Solve for \(DE\)**:

\[

DE = 18 \cdot \frac{7}{3} = 18 \cdot \frac{7}{3} = 6 \cdot 7 = 42

\]

Thus, the length of \(DE\) is \(\boxed{42}\).

bader Jul 26, 2024

#2**+1 **

Alright. I'm not the best with these problems, but let me give it a shot,

First, let's use the Law of Cosines. According to the law, we have

\(BC^2 = AB^2 + AC^2 - 2(AB* AC) cos ( BAC)\)

Now, we know a lot of the information given already, so we want to find cos BAC. Thus, we have

\(8^2=3^2+7^2-2(3*7) \cdot cos(BAC)\)

Now, isolating BAC, we have that

\(64=16 \cdot cos(BAC)\\ cos(BAC)=4\)

Now, why not apply Law of Cosines again? Let's do it! :)

This time, let's use the equation

\(DE^2 = AE^2 + AD^2 - 2(AE * AD) cos (BAC) \)

We figured out and know every term in the right side of the equation, so plugging in all the information, we have

\(DE^2 = 9^2+21^2-2(9*21)*4\)

mmm...let's check something out.

This simplifies to \(DE^2=-990 \), which defintely IS NOT possible.

Now, disclaimer though....it's kinda late at night, and my brain is fried. my solution is probably faulty.

...

NotThatSmart Jul 26, 2024

#3**+1 **

Best Answer

Using the Law of Cosines twice :

[BC^2 - BA^2 - AC^2] / [ -2 BA * AC] = cos DAE

[8^2 - 3^2 - 7^2 ] / [ -2 * 3 * 7 ] = cos DAE = -1/7

And

DE^2 = DA^2 + AE^2 - [2* DA * AE] cos ( DAE)

DE^2 = 21^2 + 9^2 - [ 2* 21 * 9] (-1/7)

DE^2 = 522 + 54

DE^2 = 576

DE = sqrt 576 = 24

CPhill Jul 26, 2024