Similar triangles - hot to determine??

[katherine.st.onge]

I took a practice test today and came across a problem in which one triangle was set inside of another. The problem said that two sides (one from each triangle) were parallel. The angle opposite this side was shared by both triangles. The side between the angle and the parallel side was twice as big on the larger triangle than the smaller.

My question is -- how do parallel sides relate to triangles being similar??


The question found on the test is below, but I was unable to paste the diagram.

C is the center of the circle, DE || CA.

Quantity A

The measure of angle DEB

Quantity B

The measure of angle DBE

Because DE || CA, angle DEB is equal to angle CAB. The comparison can therefore also be regarded as being between angles CAB and DBE. Let us focus on triangle ABC: two of its sides, namely AC and BC, are radii of the circle, and are therefore equal in length. This means that triangle ABC is isosceles. The angles opposite the equal sides must be equal: angle CAB, which is opposite side BC, is equal to angle CBA, which is opposite side AC. Note that angle CBA is literally the same angle as angle DBE:


Quantity A

Quantity B
The measure of angle DEB = the measure of angle CAB

The measure of angle DBE = the measure of angle CBA

The correct answer is C.

[jen]

Good question! This happens ALL THE TIME on the GRE -- two triangles with parallel sides turn out to be similar.

Here are some pictures:
http://www.mathwarehouse.com/geometry/s ... triangles/

Why does this work? Well, if triangles have two parallel sides, they will share two angles. In the top pic at the link above:
-- if DE and BC are parallel
-- then angle EDA is the same as angle CBA
-- and angle AED is the same as angle ACB

Any triangles that have 2 identical angles actually have all 3 angles identical, since the angles in a triangle have to sum to 180. In this case, it's even more obvious, since the third angle of each triangle is actually the SAME angle (angle a).

Sincerely,
Jen