Trapezoid Mystery

Geometry Level 3

In the diagram below, $ABGF$ is a trapezoid and $D$ is the intersection point of its diagonals.

If $CE$ is drawn such that it passes through $D$ and is parallel to $AB,$ is it true that $CD = DE?$

True False Not enough information

3 solutions

Worranat Pakornrat
May 15, 2017

With the same base & height, the triangles $AFG$ and $BFG$ have the same area. Subtracting the area of triangle $DFG$ , we can prove that the triangles $ADF$ & $BDG$ have the same area. Let $CD=x$ , $DE=y$ , and $h_1$ & $h_2$ be the heights from $AB$ to $CE$ and from $CE$ to $FG$ respectively.

Then $\dfrac{x}{2}(h_{1}+h_2)=\dfrac{y}{2}(h_{1}+h_2)$ .

Thus, $x=y$ . $CD=DE$ .

Moderator note:

This problem uses an identical trick (and nearly identical diagram) from a problem last week ; just like last week, the fastest approach is to notice the overlapping triangles.

There's a "more obvious" solution by the similarity of triangles due to the parallel lines.

$\frac{ CD}{AB} = \frac{FC}{FA} = \frac{GE}{GB} = \frac{ ED}{AB}.$

Calvin Lin Staff - 4 years ago

Oh, OK. Thank you for sharing. :)

Worranat Pakornrat - 4 years ago

@Calvin Lin -- In fact, any horizontal line parallel to AB will yield a pair of segments of equal length when cut by AF - BF and by AG - BG. Going through D is not needed.

Tom Verhoeff - 4 years ago

If you wish to kill a fly with a bazooka, we may apply an affine transformation sending $ABFG$ to an isosceles trapezoid while preserving $CD/DE$ . Clearly now we see $CD/DE=1$ so that must be true in the original problem as well!

Daniel Liu - 4 years ago

LOL, that was indeed my first impression.

I decided to figure out a more elementary explanation, which relies on those ideas.

Calvin Lin Staff - 4 years ago

Mario Piccinin
May 24, 2017

Relevant wiki: Crossed Ladders Problem

Can use 'crossed ladders theorem', whereby the height of the intersecting lines is equal to the reciprocal of the sum of the inverses of the two sides. CD = 1 / ( 1/AB + 1/FG ) similarly ED = 1 / (1/FG + 1/AB)

hence equality

Oh, I didn't expect to see this theorem appear again. Nice!

Pi Han Goh - 4 years ago

Consider the trapezium as a perspective of a playing field. The diagonals cross at the center of the field, so the line CE, being parallel to the ends is the like half-way line bisected equally by the diagonals.

David Pearce - 4 years ago

Prove congrency and heights will be equal

Chaz Dodd - 4 years ago

Oops! My bad. 😳

Chaz Dodd - 4 years ago

Harshul Trivedi
May 26, 2017

By using basic proportionate theorem and extend it for mid point , you will get the answer.. Consider this as hint

Trapezoid Mystery

3 solutions

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