+118 In quadrilateral BCDE, 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?
+1167 We can divide the problem into two triangles: △ABC and △ACE. Since we are given three side lengths for each triangle, we can use the [Law of Cosines] to solve for DE.
For △ABC, we have
\begin{align*} BC^2 &= AB^2 + AC^2 - 2 \cdot AB \cdot AC \cos ( \angle A) \ 8^2 &= 3^2 + 7^2 - 2 \cdot 3 \cdot 7 \cos ( \angle A) \ \cos ( \angle A) &= \frac{38}{42} = \dfrac{19}{21}. \end{align*}
Similarly, for △ACE, we obtain
\begin{align*} AC^2 &= AE^2 + CE^2 - 2 \cdot AE \cdot CE \cos ( \angle A) \ 7^2 &= (AE + DE)^2 + 2^2 - 2 \cdot (AE + DE) \cdot 2 \cos ( \angle A) \ 7^2 &= AE^2 + DE^2 + 4AE + 4 - 4(AE + DE) \cdot \dfrac{19}{21} \ 0 &= AE^2 + DE^2 - \dfrac{32}{7} AE - \dfrac{32}{7} DE + 20. \end{align*}
We can complete the square on both the AE2 and DE2 terms. To do this, we take half of the coefficient of our AE term, square it, and add it to both sides of the equation. Similarly, we do this for the DE term.
This gives us
\begin{align*} 0 &= (AE^2 - \dfrac{16}{7} AE + \dfrac{64}{49}) + (DE^2 - \dfrac{16}{7} DE + \dfrac{64}{49}) + 20 - \dfrac{64}{7} (AE + DE) \ &= (AE - \dfrac{8}{7})^2 + (DE - \dfrac{8}{7})^2 + 20 - \dfrac{64}{7} (AE + DE). \end{align*}
Let x=AE−78 and y=DE−78. Then AE=x+78 and DE=y+78, so
\begin{align*} 0 &= (x)^2 + (y)^2 + 20 - \dfrac{64}{7} ((x + \dfrac{8}{7}) + (y + \dfrac{8}{7})) \ 0 &= x^2 + y^2 + 20 - \dfrac{64}{7} x - \dfrac{64}{7} y - 16. \ 0 &= x^2 + y^2 - \dfrac{64}{7} x - \dfrac{64}{7} y + 4. \end{align*}
Now we can complete the square again, this time on the entire right side of the equation. Taking half of the coefficient of our x term, squaring it, and adding it to both sides gives
\begin{align*} 0 &= (x^2 - \dfrac{64}{7} x + \dfrac{16^2}{49}) + (y^2 - \dfrac{64}{7} y + \dfrac{16^2}{49}) + 4 - \dfrac{16^2}{49}. \ &= (x - \dfrac{32}{7})^2 + (y - \dfrac{32}{7})^2 - \dfrac{225}{49}. \end{align*}
Since the square of a real number is never negative, the only way for the right side of this equation to equal zero is if both (x-32/7)2 and (y−32/7)2 are zero. Thus, x = 32/7 and y = 32/7, which means AE=8 and DE=8.
Therefore, DE=8.
