From the ends point B and point C of one of the sides BC of the triangle ABC, let the two straight lines BD and DC be constructed meeting within the triangle.

I say that the sum of line BD and line DC is less than the sum of the remaining two sides of the triangle line BA and line AC, but line BD and line DC contain an angle BDC greater than the angle BAC.

Draw line BD through to point E.

Since in any triangle the sum of two sides is greater than the remaining one, therefore, in the triangle ABE, the sum of the two sides AB and AE is greater than line BE.

Add line EC to each. Then the sum of line BA and line AC is greater than the sum of line BE and line EC.

Again, since, in the triangle CED, the sum of the two sides line CE and line ED is greater than line CD, add line DB to each, therefore the sum of line CE and line EB is greater than the sum of line CD and line DB.

But the sum of line BA and line AC was proved greater than the sum of line BE and line EC, therefore the sum of line BA and line AC is much greater than the sum of line BD and line DC.

Again, since in any triangle the exterior angle is greater than the interior and opposite angle, therefore, in the triangle CDE, the exterior angle BDC is greater than the angle CED.

For the same reason, moreover, in the triangle ABE the exterior angle CEB is greater than the angle BAC. But the angle BDC was proved greater than the angle CEB, therefore the angle BDC is much greater than the angle BAC.

Therefore if from the ends of one of the sides of a triangle two straight lines are constructed meeting within the triangle, then the sum of the straight lines so constructed is less than the sum of the remaining two sides of the triangle, but the constructed straight lines contain a greater angle than the angle contained by the remaining two sides.

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