Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/339

Rh 30. ABC is a triangle, and O is the point of intersection of the perpendiculars from A, B, C on the opposite sides of the triangle: the circle which passes through the middle points of A, OB, OC will pass through the feet of the perpendiculars and through the middle points of the sides of the triangle.

Let D, E, F be the middle points of OA, OB, OC respectively; let G be the foot of the perpendicular from A on BC, and H the middle point of BC.

Then OBC is a right-angled triangle and E is the middle point of the hypotenuse OB; therefore EG is equal to EO; therefore the angle EGO is equal to the angle EOG. Similarly, the angle FGO is equal to the angle FOG. Therefore the angle FGE is equal to the angle FOE. But the angles FOE and BAG are together equal to two right angles; therefore the angles FGE and BAC are together equal to two right angles. And the angle BAC is equal to the angle EDF, because ED, DF are parallel to BA, AG (VI. 2). Therefore the angles FGE and EDF are together equal to two right angles. Hence G is on the circumference of the circle which passes through D, E, F {Note on III. 22).

Again, FH is parallel to OB, and EH parallel to 0C therefore the angle EHF is equal to the angle EGF. Therefore H is also on the circumference of the circle.