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A Perpendicular being drawn from the Right Angle of a Right Angled Tri∣angle to the opposite side, divideth the same into Two Triangles which are a like thereto.
If from the Right Angle ABC be drawn a perpendicular BD to the oppo∣site side AC, it will divide the Right Angled Triangle ABC into Two Trian∣gles ADB, BDC, which shall be like, or equiangular to the Triangle ABC.
Demonstration. The Triangles ABC, ADB, have the same Angle A; the Angle ADB, ABC, are right: they are thence equiangular (by the Cor. 2. of the 32d. of the 1st.) In like manner the Triangles BDC, ABC, have the Angle C common; and the Angles ABC, BDC, being right, they are also equal. Thence the Triangles ABC, DBC, are like.
WE measure inaccessable distances by a Square, according to this