2003 AMC 10A Problem 22

From the rectangle, $CH = BC - BH = 9 - 6 = 3$ and $EA = AD - DE = 9 - 4 = 5$.

Since $BC \parallel AD$, the lines $GA$ and $GE$ cut the two parallels in similar triangles: $\triangle GCH \sim \triangle GEA$ with ratio $\dfrac{CH}{EA} = \dfrac{3}{5}$.

Extend line $BC$ until it meets line $GF$ at a point $X$. Because $GF$ is perpendicular to $AD$, it is perpendicular to $BC$ as well, so $GX$ and $GF$ are the altitudes from $G$ of the two similar triangles, and $XF$ equals the distance between the parallel sides, $CD = 8$.

Corresponding altitudes are in the similarity ratio:
$$\frac{GX}{GF} = \frac{3}{5}, \qquad GF - GX = 8.$$
So $GF - \tfrac{3}{5} GF = 8$, giving $\tfrac{2}{5} GF = 8$ and $GF = 20$.

The answer is $\boxed{\textbf{(B)}\ 20}$.

Two lines through a common point crossing a pair of parallels always produce similar triangles, and every corresponding length, including altitudes, scales by the same ratio. The perpendicular $GF$ is that altitude, and the rectangle supplies the fixed gap of $8$ between the two altitudes.

Coordinates with $D = (0,0)$, $A = (9,0)$, $C = (0,8)$, $B = (9,8)$: then $E = (4,0)$ and $H = (3,8)$. Line $EC$: $y = -2x + 8$. Line $AH$: $y = -\tfrac{4}{3}(x - 9)$. Solving, $-2x + 8 = -\tfrac43 x + 12$ gives $x = -6$, $y = 20$. So $G = (-6, 20)$ and $GF$ is its height above line $AD$, namely $20$.

• Using BH = 6 and DE = 4 as the similarity ratio instead of the correct pieces CH = 3 and EA = 5, or forgetting to add the 8 back to the smaller altitude.
• Pairing the wrong triangles (for instance assuming $\triangle GFA \sim \triangle ABH$ without also using a second relation), which leaves one unknown too many.