IN any Triangle whatever, the Square of the side opposite to an acute Angle, together with two Rectangles comprehend∣ed under the side on which the Perpendi∣cular falleth, and under the Line which is betwixt the Perpendicular and that An∣gle, is equal to the Square of the other sides.
Let the proposed Triangle be ABC, which hath the Angle C, acute; and if one draw AD perpendicular to BC, the Square of the side AB, which is opposite to the acute Angle C, toge∣ther with two Rectangles comprehend∣ed under BC, DC, shall be equal to the Squares of AC, BC.
Demonstration. The Line BC is di∣vided in D; whence (by the 7th.) the Square of BC, DC, are equal to two Rectangles under BC, DC, and to