1. Test Question

A friend and I took a test yesterday for our geometry and discrete math course. The question that puzzled us is:
in triangle ABC, BC is produced to E and CB is produced to D. AC = AB and the ratio of sides is: AB : DB as CE:AB. (or AB/DB = CE/AB). PRove triangle ABD is similar to ACE.

Now our class basically all wrote they were similar and congruent by somehow proving that DB = CE. my friend and i wrote the same because thats what the teacher was expecting.
You can provethey are similar through SAS proof, IF you assume DB = CE. But how do you prove they are the same? We think you cant, andtherefore that ratio given does not work for all values that you can input for AB, AC, DB and CE. for example;

if AB = AC and AB = 6, and DB = 4. When you plug in these values into the ratio and isolate for CE, you get CE is 9. therefore CE does not equal DB.

our question is are we right? or we missing something really obivous here.

the answer the teacher said after the test was that the two triangles were a special case of similarity -> congruent. thank you in advance

2. I agree with you they are not.

Because the two outer triangles has different area. Since the area of the triangle is half product of the sides and the sine of the angle they make. But those angles must be the same because we have an isoseles triangle. So there is no way they are congruent.

3. Hello, Finroe!

As TPHacker pointed out, the two triangles are not congruent . . .

In triangle ABC, BC is produced to E and CB is produced to D.
AC = AB and the ratio of sides is: $\frac{AB}{DB} \,= \,\frac{CE}{AB}$
Prove triangle ABD is similar to ACE.
Code:
                    A
*
*/ \  *
* /   \     *
*  /     \        *
*   /       \           *
*    /         \              *
*     /           \                 *
*      /             \                    *
* - - - * - - - - - - - * - - - - - - - - - - - *
D       B               C                       E

From $\frac{AB}{DB} \,= \,\frac{CE}{AB}$ and $AB = AC$, we have: . $\frac{AB}{DB} \,= \,\frac{CE}{AC}$

Since $\angle ABC$ and $\angle ACB$ are base angles of an isosceles triangle: . $\angle ABC\,=\,\angle ACB$

. . Hence: . $\angle ABD \,= \,\angle ACE$ . (Supplements of equal angles are equal.)

Therefore, by "sas": . $\Delta ABD \sim \Delta ACE$

4. Originally Posted by Finroe
A friend and I took a test yesterday for our geometry and discrete math course. The question that puzzled us is:
in triangle ABC, BC is produced to E and CB is produced to D. AC = AB and the ratio of sides is: AB : DB as CE:AB. (or AB/DB = CE/AB). PRove triangle ABD is similar to ACE.
Are you sure that you don't have a typo?

It looks like it should read : AB : DB as AB:CE.

RonL

5. Originally Posted by CaptainBlack
Are you sure that you don't have a typo?

It looks like it should read : AB : DB as AB:CE.

RonL
Yeh theres no typo. I clarified with my teacher later on today. You guys are right, the triangles ARE similar, but NOT congruent. Thanks for all the help.