In rectangle ABCD below, points F and G lie on segment AB such that AF = FG = GB and E is the midpoint of segment DC. Also, segment AC intersects segment EF at H and segment EG at J. The area of rectangle ABCD is 70. Find the area of triangle AHF.

(Note: This question has been slightly changed from the original AMC 12 problem.)



  • Triangle AHF is similar to triangle CHE with ratio 2:3
  • Triangle AJG is similar to triangle CJE with ratio 4:3
  • $\begingroup$ Don't we need $J$ and $G$? or Is the $\Delta AHF$ this one? $\endgroup$ Apr 26, 2015 at 10:46
  • 1
    $\begingroup$ Did the joker do the drawing? $\endgroup$
    – Jiminion
    Nov 14, 2017 at 20:07

2 Answers 2


Let's say that $AB=3x$, $CD=2y$ and $BC=h$. Then, $$3xh = 2yh = 70,$$and $$x = \frac{2y}{3}.$$ The triangles $AFH$ and $HEC$ are similar and have area:

$$S_{AFH} = \frac{x h_1}{2}, S_{HEC} = \frac{y h_2}{2}$$

where $h_1$ and $h_2$ are the heights of the 2 triangles, with $h_1+h_2 = h$ and $h_1 = \frac{2h_2}{3}$. Then: $$h_1 = \frac{2h}{5}, h_2 = \frac{3h}{5},$$ and since $xh = \frac{70}{3}$, then:

$$S_{AFH} = \frac{x h}{5} = \frac{70}{15} = \frac{14}{3}$$


There's a "straightforward" vector solution:
Let $AB=b$, $AD=d$ the basis vectors and $[x\times y]$ be the cross product, since we're given $|[AB\times AD]|= 70$.
Vectors $AF=1/3AB=1/3b, AG=2/3AB=2/3b, AE=AD+1/2DE=AD+1/2AB=d+1/2b$.
$X$ lies on the line $YZ$ iff $AX=t\cdot AY + (1-t)\cdot AZ$, where $t$ is a real.
So, consider $AH=(1-u)\cdot 0+u\cdot (b+d)=(1-v)\cdot AE+v\cdot AF$
$u\cdot (b+d)=(1-v)\cdot (d+1/2b)+v/3\cdot b$ $u=(1-v)/2+v/3, u=(1-v) \Rightarrow u=2/5, v=3/5 \Rightarrow AH=2/5\cdot(b+d)$

$S_{\Delta AHF} = \frac{1}{2}[AH\times AF]= \frac{1}{2}[2/5\cdot(b+d)\times 1/3b]= \frac{1}{15}[(b+d)\times b] = \frac{1}{15}\left([b\times b] + [d\times b] \right)= \frac{1}{15} (0+70) = \frac{14}{3}$. More classical way:

Triangle $AHF$ is similar to triangle $CHE$ with ratio $2:3$

That is all one needs. $AH/HC=2/3$, so $AH/AC=2/5$ and the "height" of $\Delta AHF$ is $2/5AD$, while its base is $1/3AB$ , $S=\frac{1}{2}\cdot\frac{2}{5}\cdot\frac{1}{3}\cdot 70=\frac{14}{3}$

  • $\begingroup$ Thanks! Both answers were helpful, but this came first and so it gets the check. $\endgroup$
    – Rex
    Apr 26, 2015 at 12:51

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