In any triangle the sum of any two sides is greater than the remaining one.

Let *ABC* be a triangle.

I say that in the triangle *ABC* the sum of any two sides is greater than the remaining one, that is, the sum of *BA* and *AC* is greater than *BC,* the sum of *AB* and *BC* is greater than *AC,* and the sum of *BC* and *CA* is greater than *AB.*

Draw *BA* through to the point *D,* and make *DA* equal to *CA.*

Join *DC.*

Since *DA* equals *AC,* therefore the angle *ADC* also equals the angle *ACD.* Therefore the angle *BCD* is greater than the angle *ADC.*

Since *DCB* is a triangle having the angle *BCD* greater than the angle *BDC,* and the side opposite the greater angle is greater, therefore *DB* is greater than *BC.*

But *DA* equals *AC,* therefore the sum of *BA* and *AC* is greater than *BC.*

Similarly we can prove that the sum of *AB* and *BC* is also greater than *CA,* and the sum of *BC* and *CA* is greater than *AB.*

Therefore *in any triangle the sum of any two sides is greater than the remaining one.*

Q.E.D.

Suppose there are two points A and B on the same side of a line CD. The problem is to find the shortest path which goes first from the point A to some point P on the line CD, then from P to the point B. We will only consider paths that are made out of straight lines; call such a path a bent line. But that still leaves us the question of which point P to choose on the line CD to minimize the sum of the distances AP plus PB.
The solution is that the shortest path will be the path
First, we should show how to construct the bent line where the angle of incidence equals the angle of reflection. Draw a perpendicular |

Now, triangles *BFE* and *B'FE* are congruent since they have two sides and the included angle equal (I.4), the included angles being right angles. Therefore, angles *BFE* and *BF'E* are equal. The vertical angle *AEC* across from angle *B'ED* also equals these angles (I.15). Thus, the angle of incidence *AEC* equals the angle of reflection *BED.*

We still have to show that the distance *AE* + *EB* is less than any distance *AP* + *EP* for any point *P* other than *E* that lies on the line *CD.* Let *P* be such a point and draw lines *AP, BP,* and *B'P.* Then by proposition I.20, above, *AP* + *EP* is less than *AB'.* But *AB'* = *AE* + *EB',* and *EB'* = *EB,*
therefore *AP* + *EP* is less than *AE* + *EB.*

Thus, the shortest bent line between two points on the same side of a line that meets that line is the one where the angle of incidence equals the angle of reflection.

Q.E.D.