# Proving Side-Side-Side congruency of a triangle

• Sep 14th 2008, 10:59 AM
reagan3nc
Proving Side-Side-Side congruency of a triangle
I need to prove SSS traingle congruency using Side-Angle-Side as an axiom. I know that if 3 sides of a triangle are congruent to 3 sides of another triangle the 2 triangles are equivalent. I am just not sure how to write up in a proof using SAS as the axiom or the given. Thanks for any help anyone can give. In the question it said to use a kite or dart to help with the proof, but I do not see how that can help.
• Sep 14th 2008, 12:25 PM
Soroban
Hello, reagan3nc!

Quote:

I need to prove SSS triangle congruency using SAS as an axiom.

It said to use a kite or dart to help with the proof

Code:

                      B                       *                   * α |β*               *      |  *           *          |    *     A * - - - - - - - * - - - * C           *          |E    *               *      |  *                   * α |β*                       *                       D

We are given two triangles with equal corresponding sides.
Place them so they have an equal side in common.

We have $\displaystyle \Delta ABC$ and $\displaystyle \Delta ADC$ with: .$\displaystyle AB = AD,\;BC = CD$
. . and, of course, $\displaystyle AC = AC.$

Since $\displaystyle AB = AD,\;\Delta ABD$ is isosceles.
Hence, its base angles are equal: .$\displaystyle \angle ABD = \angle ADB = \alpha$

Since $\displaystyle BC = CD,\;\Delta BCD$ is isosceles.
Hence, its base angles are equal: .$\displaystyle \angle CBD = \angle CDB = \beta$

Since $\displaystyle \begin{Bmatrix}\angle ABC &=& \alpha + \beta \\\angle ACD &=& \alpha + \beta\end{Bmatrix}$ . then: .$\displaystyle \angle ABC = \angle ADC$

And we have: . $\displaystyle \begin{array}{ccc}\Delta ABC & &\Delta ADC \\ \hline AB &=& AD \\ \angle ABC &=& \angle ADC \\ BC &=& CD \end{array}$

Therefore: . $\displaystyle \Delta ABC \cong \Delta ADC\;\text{ by SAS}$