1. ## Proving Triangle Sides

Hey everyone,

I have a question here that states:

Suppose AB=DE and BC=EF for triangle ABC and triangle DEF, but angle B is greater than angle E. Then AC is greater than DF.

I can quite easily draw the diagrams and understand it by looking at it, but I am stumped on how to put this one into words. Any help would be greatly appreciated.

Thanks guys

2. For a triangle with sides a,b,c, use the cosine formula $c^2 = a^2 + b^2 - 2ab \cos(\gamma)$

since on both triangles you have, a=d and b=e, then you just have to look at the $- 2ab \cos(\gamma)$.

Since $\cos(\gamma)$ is a decreasing function on $[0,\pi]$, if $\gamma > x$ where x is an angle (and they're both contained in $[0,\pi]$) then $\cos(\gamma) < \cos(x)$.

So for your triangles, the triangle with the larger angle will result in a smaller cos value and hence $c^2$ will be larger.

I don't think this is clear what I've written so reply back if not...

3. Originally Posted by falloutboy10
Hey everyone,

I have a question here that states:

Suppose AB=DE and BC=EF for triangle ABC and triangle DEF, but angle B is greater than angle E. Then AC is greater than DF.

I can quite easily draw the diagrams and understand it by looking at it, but I am stumped on how to put this one into words. Any help would be greatly appreciated.

Thanks guys
Here is another view,

in conjunction with the Cosine Rule as outlined by Deadstar,
showing the angle in triangle ABC at E < the angle at B in triangle DEF

Then, since $AB=DE,\ BC=EF$

$AC^2=AB^2+BC^2-2(AB)(BC)Cos(angle B)$

$DF^2=AB^2+BC^2-2(AB)(BC)Cos(angle E)$

the difference is $AC^2-DF^2=2(AB)(BC)[Cos(angle E)-Cos(angle B)]$

The angles are < 180 degrees,
Cosine gives the x co-ordinate which starts at R, the radius, for an angle 0,
decreasing through zero all the way to -R for an angle of 180 degrees.

Hence from 0 to 180 degrees...

Cos(angle) $>$ Cos(greater angle)

Hence the difference calculated above is positive.
this means AC $>$ DF