23. 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;

E

D

B

D

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 (III. 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 BCG the perpendiculars BF and CE meet at A; therefore by the former case the straight line GA produced will be perpendicular to BC.

29. If from any point in the circumference of the circle described round a triangle perpendiculars be drawn to the sides of the triangle, the three points of intersection are in the same straight line.

Let ABC be a triangle, P any point on the circumference of the circumscribing circle; from P draw PD,

C

Ε

PE, PF perpendiculars to the sides BC, CA, AB respectively: D, E, F shall be in the same straight line.

[We will suppose that P is on the arc cut off by AB, on the opposite side from C, and that E is on CA produced through A; the demonstration will only have to be slightly modified for any other figure.]

A circle will go round PEAF (Note on III. 22); therefore the angle PFE is equal to the angle PAE (III. 21). But the angles PAE and PAC are together equal to two right angles (I. 13); and the angles PAC and PBC are together equal to two right angles (III. 22). Therefore the angle PAE is equal to the angle PBC; therefore the angle PFE is equal to the angle PBC.

Again, a circle will go round PFDB (Note on III. 21); therefore the angles PFD and PBD are together equal to two right angles (III. 22). But the angle PBD has been shewn equal to the angle PFE. Therefore the angles PFD and PFE are together equal to two right angles. Therefore EF and FD are in the same straight line.

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 OA, 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.

H

Then OBG 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 FÕE and BAC 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, AC (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 OC; therefore the angle EHF is equal to the angle EGF. Therefore H is also on the circumference of the circle.

Similarly, the two points in each of the other sides of the triangle ABC may be shewn to be on the circumference of the circle.

The circle which is thus shewn to pass through these nine points may be called the Nine points circle: it has some curious properties, of which we will now give two.

The radius of the Nine points circle is half of the radius of the circle described round the original triangle.

For the triangle DEF has its sides respectively halves of the sides of the triangle ABC, so that the triangles are similar. Hence the radius of the circle described round DEF is half of the radius of the circle described round ABC.

If S be the centre of the circle described round the triangle ABC, the centre of the Nine points circle is the middle point of SO.

For HS is at right angles to BC, and therefore parallel to GO. Hence the straight line which bisects HG at right angles must bisect SO. And Hand G are on the circumference of the Nine points circle, so that the straight line which bisects HG at right angles must pass through the centre of the Nine points circle. Similarly, from the other sides of the triangle ABC two other straight lines can be obtained, which pass through the centre of the Nine points circle and also bisect SO. Hence the centre of the Nine points circle must coincide with the middle point of SO.

We may state that the Nine points circle of any triangle touches the inscribed circle and the escribed circles of the triangle: a demonstration of this theorem will be found in the Plane Trigonometry, Chapter XXIV. For the history of the theorem see the Nouvelles Annales de Mathématiques for 1863, page 562.

31. If two straight lines bisecting two angles of a triangle and terminated at the opposite sides be equal, the bisected angles shall be equal.

Let ABC be a triangle; let the straight line BD bisect the angle at B, and be terminated at the side AC; and let the straight line CE bisect the angle at C, and be terminated at the side AB; and let the straight line BD be equal to the straight line CE: then the angle at B shall be equal to the angle at C.

E

For, let BD and CE meet at O; then if the angle OBC be not equal to the angle OCB, one of them must be greater than the other; let the angle OBC be the greater. Then, because CB and BD are equal to BC and CE, each to each; but the angle CBD is greater than the angle BCE; therefore CD is greater than BE (I. 24).

On the other side of the base BC make the triangle BCF equal to the triangle CBE, so that BF may be equal to CE, and CF equal to BE (I. 22); and join DĚ.

Then because BF is equal to BD, the angle BFD is equal to the angle BDF. And the angle OCD is, by hypothesis, less than the angle OBE; and the angle COD is equal to the angle BOE; therefore the angle ODC is greater than the angle OEB (I. 32), and therefore the angle ODC is greater than the angle BFC.

Hence, by taking away the equal angles BDF and BFD, the angle FDC is greater than the angle DFC; and therefore CF is greater than CD (I. 19); therefore BE is greater than CD.

But it was shewn that CD is greater than BE; which is absurd.

Therefore the angles OBC and OCB are not unequal, that is, they are equal; and therefore the angle ABC is equal to the angle ACB.

[For the history of this theorem see Lady's and Gentleman's Diary for 1859, page 88.]