43. The bisector of an angle of a triangle

The lesson, step by step

1. Warm up

   Ask: State and prove Prop. XIII (the parallel to a side).

2. The new idea

   The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the adjacent sides.

   The bisector of an exterior angle of a triangle divides the opposite side externally into segments which are proportional to the adjacent sides.

3. Draw the figure

   Draw triangle CAB with CA about 6 cm and CB about 10 cm. Bisect angle C with the compasses; the bisector meets AB at M. Through A draw AE parallel to MC, meeting BC extended at E.

4. Walk through the proof

   XV: by Prop. XIII, MA : MB = CE : CB. The angles at A and E equal the two halves of angle C (alternate-interior and exterior-interior angles), so triangle ACE is isosceles: CE = CA. Substitute.

   XVI: the same idea with the exterior angle, drawing AF parallel to the bisector.

5. Your student proves it back

   Close the book. Your student redraws the figure from memory, letters it the same way, and gives the proof aloud: what is given, what is to be proved, then each step with its reason ("because the radii are equal", "by § 143"). Point to each line or angle as it is named.

   If a step is missing, ask "How do we know that?" rather than supplying it.

6. Exercises

   Ex. 249. In a triangle ABC, AB = 12, AC = 14, BC = 13. Find the segments of BC made by the bisector of the angle A.

   Ex. 250. In a triangle CAB, CA = 6, CB = 12, AB = 15. Find the segments of AB made by the bisector of the angle C.

7. Wrap up

   Measure MA and MB in your drawing and compare their ratio with CA : CB.