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A high-schooler I know had this problem for homework:

Given:

  • trapezoid $ABCD$ with $\overline{AB}\Vert\overline{CD}$ and $m\angle C=90^\circ$;
  • $E$ is the midpoint of $\overline{AD}$;
  • $\overline{CB}$ is extended outside the trapezoid to $F$ such that $\overline{FE}\perp\overline{AD}$;
  • $CD=12$; $AB=18$; $CB=8$.*

Find the length of $\overline{BF}$.

My solution:

Drop a segment from $E$ to hit $\overline{AB}$ perpendicularly at $G$. Denote by $H$ the intersection of $\overline{AB}$ and $\overline{EF}$.
(A diagram.)
Then $EG=4$ and, by Pythagoras, $AD=10$ so that $AE=5$. By Pythagoras again, $AG=3$. (You can obtain $AG=3$ from $\triangle AGE\sim\triangle A?D$ instead.) By $\triangle AGE\sim\triangle EGH$, we obtain $GH$. We then obtain $HB=AB-AH$; and, by $\triangle AGE\sim\triangle FBH$, we find $BF$.

I'm not happy with this solution; I suspect there's a much simpler one I'm missing. Does anyone see one?

* I don't recall the problem exactly, actually. It gave either $CB=8$ or $AD=10$.

msh210
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2 Answers2

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You may also consider the projection of $E$ on $CF$, say $P$. Then $EPF$ and $EGA$ are similar ($AEFB$ is a cyclic quadrilateral). Since $\frac{EG}{GA}=\frac{8}{18-12}=\frac{4}{3}$, we have $PF=\frac{3}{4}EP = \frac{3}{8}(12+18) = \frac{45}{4}$. On the other hand $PB=\frac{1}{2}CB=4$, hence by difference $BF=\frac{29}{4}$.

Jack D'Aurizio
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Where would $F$ be located on $BC$ if $E$ were at $D$? Well, we know that the projection of $D$ onto $AB$ at, say, $Q$, makes $\triangle AQD$ an $6$-$8$-$10$ Pythagorean triangle. Thus, if $E$ were at $D$, $\triangle FCD$ would be similar to $\triangle AQD$ and since $CD = 12$, we would find $FC = 6(12)/8 = 9$.

Now, what about the other extreme; i.e., if $E$ were at $A$, where would $F$ be on $BC$? Again, we have similarity, namely $\triangle FBA \sim \triangle AQD$, hence $FB = 6(18)/8 = 27/2$, consequently $FC = 8 + 27/2 = 43/2$.

But $E$ is actually located halfway between $A$ and $D$. Thus the true distance of $FC$ is halfway between $9$ and $43/2$, or $FC = 61/4$, so $BF = 29/4$.

We could have performed the same reasoning by drawing the line $EP$ perpendicular to $BC$ (as in Jack's answer), and noting that $EP = (12+18)/2 = 15$, and $\triangle FPE \sim \triangle AQD$, hence $$FP = 6(15)/8 = 45/4,$$ and since $P$ is the midpoint of $BC$, $BF = 45/4 - 4 = 29/4$.

heropup
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