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A Streight Line drawn from one point of the Circumference of a Circle to another, shall fall within the same.
Let there be drawn from the Point B, in the Circumference, a Line to the Point C. I say that it shall fall whol∣ly within the Circle. To prove that it cannot fall without the Circle, as BVC; having found the Center there∣of, which is A, draw the Lines AB, AC, AV.
Demonstration. The Sides AB, AC, of the Triangle ABC, are equal: whence (by the 5th. of the 1st.) the An∣gles ABC, ACB, are equal. And seeing the Angle AVC, is exteriour in respect of the Triangle AVB, it is greater than ABC, (by the 16th. of the 1st.) it shall be also greater than the Angle ACB. Thence (by the 19th of the 1st.) in the Triangle ACV, the side AC, opposite to the greatest An∣gle AVC, is greater than AV; and by consequence, AV cannot reach the circumference of the Circle, seeing it