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THe greatest Side of every Triangle, subtends the greatest Angle.
Let the Side BC, of the Triangle ABC, be greater then the Side AC. I say that the Angle BAC, which is opposite to the Side BC, is greater than the Angle B, which is opposite to the Side AC. Cut off from BC, the Line CD, equal to AC and draw AD.
Demonstration. Seeing that the Sides AC, CD, are equal, the Tri∣angle ACB is an Isosceles Triangle, (by the 5th.) the Angles CDA, CAD, are therefore equal. Now the whole Angle BAC is greater than the Angle CAD: Thence the Angle BAC, is greater than the Angle CDA; which being exteriour, in regard of the Tri∣angle ABD, is greater than the inte∣riour B (by the 16th.) Therefore the Angle BAC, is greater than the Angle B.