A triangle with two equal sides has two equal angles

The first payoff of SAS — proving a fact about angles by building a matching pair of triangles, not by measuring.

Two lessons ago, congruence was about comparing two separate triangles. This lesson uses it on a single triangle, split into two — the first time this course proves a fact about angles by building a matching pair of triangles inside one shape, rather than by measuring anything directly.

A triangle, A, B and C. B and C are fixed. A is draggable, but only up and down: it always sits directly above the midpoint of BC, so AB and AC stay equal to each other, however high or low A goes.

They are not forced to look equal by the drawing — the angle at each corner is read independently, straight from the geometry, and it still always agrees with the other.

A triangle with (at least) two equal sides is called isosceles. The two equal sides are its legs, and the third side is its base. This lesson’s claim is that an isosceles triangle’s base angles — the two angles that sit opposite the two equal legs, at either end of the base — are always equal to each other too.

Splitting the triangle in two to prove it

Drop a line from A straight down to M, the midpoint of BC. That line splits the one triangle into two: ABM and ACM. Compare them: AB equals AC, because the triangle is isosceles, by construction. BM equals CM, because M is the midpoint of BC. AM equals AM — the same segment, shared by both triangles. Three matching pairs of sides: SSS, agreed in the last lesson, so triangle ABM is congruent to triangle ACM.

Congruent triangles have every corresponding part equal, angles included, not only the sides that were used to prove it. The angle at B in triangle ABM is the base angle at B in the original triangle; the angle at C in triangle ACM is the base angle at C. Congruence forces those two to match — which is exactly the claim.

This is a genuinely different kind of proof from anything earlier in this course. Every triangle lesson before this one measured something directly, or built one auxiliary line and read an angle off it. This proof does not read an angle off anything: it builds two whole triangles out of one, shows they must be congruent, and lets congruence hand over the equal angles as a consequence, without ever measuring them against each other. That move — split a shape in two, prove the two halves congruent, and collect whatever else congruence guarantees — is one of the most useful moves in all of geometry, and this is where it is used for the first time.

A triangle has a base angle of 70 degrees. What is its other base angle?

Also 70 degrees, provided the triangle is isosceles with that side as its base — the two base angles are always equal to each other, whatever the actual value happens to be.

Does this proof work if A is dragged to sit directly above B instead of above the midpoint of BC?

No — moving A there breaks the AB = AC assumption the whole proof rests on, since A would no longer be equidistant from B and C. The base angles would not be forced equal any more, because the triangle would no longer be isosceles.

A different isosceles triangle has its apex A dragged so AB = AC still holds, but a construction line is dropped from B instead of A. Would splitting it that way still give two triangles provably congruent by SSS? Why or why not?

Not directly — a line from B does not land on the midpoint of anything already known to be equal on both resulting triangles, so there is no ready-made third matching pair. The construction has to start from the vertex where the two equal sides meet, not from an arbitrary vertex, for SSS to fall out this easily.

Do the exercise

This lesson counts as done once you get its exercise right. Nothing to tick off by hand.