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# Geometry Help - 2 of 2

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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

Jul 26, 2024

#3
+129806
+1

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

Jul 26, 2024

#1
+1749
-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 $$\boxed{42}$$.

Jul 26, 2024
#2
+1655
+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.

...

Jul 26, 2024
edited by NotThatSmart  Jul 26, 2024
edited by NotThatSmart  Jul 26, 2024
#3
+129806
+1

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