Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/335

Rh CDG we can shew that BG is equal to CG; and from the triangles CEG and AEG we can shew that CG is equal to AG; therefore BG is equal to AG. Then if we draw a straight line from G to the middle point of AB we can show that this straight line is at right angles to AB: that is, the line which bisects AB at right angles passes through G.

25. The straight lines drawn from the angles of a triangle to the points of bisection of the opposite sides meet at the same point.

Let ABC be a triangle; bisect BC at D, bisect CA at E, and bisect AB at F; join BE and CF meeting at G; {page contains image}} join AG and GD: then AG and GD shall lie in a straight line.

The triangle BEA is equal to the triangle BEC, and the triangle GEA is equal to the triangle GEC (I. 38); therefore, by the third Axiom, the triangle BGA is equal to the triangle BGC.

Similarly, the triangle CGA is equal to the triangle CGB. Therefore the triangle BGA is equal to the triangle CGA. And the triangle BGB is equal to the triangle CGB (1. 38); therefore the triangles BGA and BGD together are equal to the triangles CGA and CGD together. Therefore the. triangles BGA and BGD together are equal to half the. triangle ABC. Therefore G must fall on the straight line AB; that is, AG and GD lie in a straight line.