28. The perpendiculars drawn from the angles of a triangle on the opposite sides meet at the same point.
Let ABC be a triangle; and first suppose that it is not obtuse angled. From B draw BE perpendicular to CA;
from C draw CF perpendicular to AB; let these perpendiculars meet at G; join AG, and produce it to meet BC at D: then AD shall be perpendicular to BC.
For a circle will go round AEGF {Note on III. 22); therefore the angle FAG is equal to the angle FEG (III. 21). And a circle will go round BCEF (IV. 31, Note on III. 21); therefore the angle FEB is equal to the angle FCB. Therefore the angle BAD is equal to the angle BCF. And the angle at B is common to the two triangles BAD and BCF. Therefore the third angle BDA is equal to the third angle BFC (Note on I. 32). But the angle BFC is a right angle, by construction; therefore the angle BDA is a right angle.
In the same way the theorem may be demonstrated when the triangle is obtuse angled. Or this case may be deduced from what has been already shewn. For suppose the angle at A obtuse, and let the perpendicular from B on the opposite side meet that side produced at E, and let the perpendicular from C on the opposite side meet that side produced at F; and let BE and CF be produced to meet at G. Then in the triangle BGG the perpendiculars BF and CE meet at A; therefore by the former case the straight line GA produced will be perpendicular to BC.
