• August 18th 2012, 01:34 PM
Schuyla
Chuck used a dynamic geometry utility and kept drawing bow tie shapes. To do this, he used two parallel lines where A, and B are on one line, C and D are on the other. He moved only point D to change the shape. Elizabeth noted that no matter where he moved D, one thing remained true about the triangles making the bow tie. What did she observe about triangles ACE and DBE?
Attachment 24541
I don't know how to go about finding the solution. Can someone help me out please.
• August 18th 2012, 03:54 PM
HallsofIvy
I can't imagine how she could observe anything about "triangles ACE and DBE" because there is NO point "E" in your picture or in your description of the problem.
• August 18th 2012, 04:10 PM
Soroban
Hello, Schuyla!

Quote:

Chuck used a dynamic geometry utility and kept drawing bowtie shapes.
To do this, he used two parallel lines where A and B are on one line, C and D are on the other.
He moved only point D to change the shape.
Elizabeth noted that no matter where he moved D, one thing remained true about the triangles making the bowtie.
What did she observe about triangles ACE and DBE?

Code:

```              C                D     - - - - - o - - - - - - - - o - -             *|:*          *::*|             *:|:::* E  *:::::* |           *::|h::::o::::::::*  |h           *:::| *    *:::::*  |         *::* |        *::*    |     - - o - - + - - - - - o - - + - -         A    F          B    G```
Draw altitudes $CF$ and $DG.$
Since $AB \parallel CD,\;CF = DG = h.$

$\Delta ACB$ has base $AB$ and height $h.$
$\Delta ADB$ has base $AB$ and height $h.$
. . Hence, referring to areas: . $\Delta ACB \,=\,\Delta ADB.$

They both have $\Delta AEB$ in common.

Hence: . $\Delta ACB - \Delta AEB \;=\;\Delta ADB - \Delta AEB$

Therefore: . . . . . $\Delta ACE \;=\;\Delta DBE$

The two halves of the bowtie have equal areas.