The converse
Two equal base angles force two equal sides — proved separately from the last lesson, not assumed from it.
The last lesson proved that two equal sides force two equal base angles. This lesson asks the reverse question: if the two base angles happen to be equal, are the two sides forced to be equal too? A claim being true has never guaranteed its converse is — that lesson was G2.4 — so this needs its own proof.
A triangle, A, B and C. B and C are fixed. A is entirely free.
A triangle, A, B and C, where B and C are fixed and A is entirely free to drag anywhere. Most positions give unequal base angles and unequal sides together. Search for the one line where the base angles come out equal, and see what happens to AB and AC at exactly that spot. This needs JavaScript switched on.
No such spot exists — every position that gives equal base angles also gives equal sides, without exception. This lesson’s claim is the converse of the last lesson’s: swap “two equal sides” and “two equal base angles” between the assumption and the conclusion, and each becomes the other’s converse. Both directions happen to be true here, which is not guaranteed — G2.4 already showed a converse that only holds because of a real, separate argument, not because the original statement ran backwards for free.
Proving it starts by dropping the angle bisector from A down to BC, meeting it at a point M. Compare triangle ABM with triangle ACM: angle ABM equals angle ACM, because the two base angles are equal, by assumption this time, not by construction. Angle BAM equals angle CAM, because AM was built as the bisector of angle A. AM equals AM — shared. Two matching angles and the side between them: ASA, agreed alongside SSS two lessons ago, so triangle ABM is congruent to triangle ACM.
Congruence hands over every corresponding part, sides included. AB in triangle ABM corresponds to AC in triangle ACM, so they are forced equal — which is the claim. What makes this proof worth separating from the last lesson’s is that it reaches for a genuinely different pair of matching parts. G4.3 built its two triangles from SSS: two equal sides plus a shared one. This lesson builds the same two triangles from ASA: two equal angles plus the side between them. The claims sound like mirror images of each other, but the proofs are not mirror images — one leans on SSS, the other on ASA — and noticing that is more useful than treating the converse as automatic.
Isosceles and equilateral triangles, and the perpendicular and angle bisectors this pair of lessons has used as tools, reappear through the rest of this course: the perpendicular bisector by name in the next lesson, the angle bisector by name in the one after that.
A triangle has base angles of 55 and 55 degrees. What can be said about its sides?
The two sides opposite those angles are equal — this lesson’s claim, not an assumption. The triangle is isosceles.
A triangle has base angles of 55 and 56 degrees. Are the two sides opposite them equal?
No, not exactly — this lesson’s proof needs the angles to actually be equal, not merely close. Sides that are very nearly, but not exactly, equal correspond to angles that are very nearly, but not exactly, equal too; there is no threshold where “close enough” starts counting.
If a third proof tried to use SAS instead, what two sides and included angle would it need, and does the given information supply them?
It would need two equal sides with the angle between them already known equal — but this lesson only starts with two equal angles, not two equal sides, so the sides SAS would require are exactly the thing still to be proved. SAS cannot get off the ground here; that is why the proof reaches for ASA instead.
This lesson counts as done once you get its exercise right. Nothing to tick off by hand.