26. The straight lines which bisect the angles of a triangle meet at the same point.
Let ABC be a triangle; bisect the angles at B and C
by straight lines meeting at G; join AG: then AG shall bisect the angle at A.
From G draw GD perpendicular to BC, GE perpendicular to CA, and GF perpendicular to AB.
From the triangles BGF and BGD we can shew that GF is equal to GD, and from the triangles CGE and CGD we can shew that GE is equal to GD; therefore GF is equal to GE. Then from the triangles AFG and AEG we can shew that the angle FAG is equal to the angle EAG.
The theorem may also be demonstrated thus. Produce AG to meet BC at H. Then AB is to BH as AG is to GH, and AC is to CH as AG is to GH (VI. 3); therefore AB is to BH as AC is to CH(V.11); therefore AB is to AC as BH is to CH (V. 16); therefore the straight line AH bisects the angle at A (VI. 3).
27. Let two sides of a triangle he produced through the base; then the straight lines which bisect the two exterior angles thus formed, and the straight line which bisects the vertical angle of the triangle, meet at the same point.
This may be shewn like 26: if we adopt the second method we shall have to use VI. A.
