The converse of Pythagoras

If the squares add up, the angle is a right angle — proved by congruence, not by area.

The last lesson started from a right angle and found a relationship between the squares. This lesson runs it the other way: a triangle, A, B and C, all three points draggable, with nothing pre-built to already be a right triangle.

A triangle, A, B and C, all three points draggable, nothing pre-built to already be a right triangle. A live reading finds the longest side, adds the squares of the other two, and reads the angle opposite the longest side — a genuine right angle exactly when the two squares match. This needs JavaScript switched on.

Drag any point. The angle opposite the longest side is 90 degrees exactly when its square matches the sum of the other two.

Unlike the last lesson, nothing here starts out as a right triangle — you have to find the one arrangement, out of infinitely many, where it becomes one. The vocabulary needed for that search is already in place: the last lesson introduced the hypotenuse, and nothing new needs naming here. What’s new is the direction of the claim itself — this lesson starts from the side lengths and concludes something about an angle, the reverse of what the theorem does.

Borrowing a right angle by congruence

Say a triangle has sides $a$ , $b$ and $c$ , with $c$ the longest, and $a^2 + b^2 = c^2$ . Build a second, separate triangle: a genuine right angle between two legs of length $a$ and $b$ . By the last lesson, that right triangle’s hypotenuse is $\sqrt{a^2+b^2} = \sqrt{c^2} = c$ .

The two triangles now have all three sides equal: $a$ , $b$ and $c$ , matched in both. By SSS, they are congruent — the same fact this course used to prove isosceles triangles, several chapters ago. Congruent triangles carry every angle over unchanged, so the angle between the sides of length $a$ and $b$ in the original triangle must equal the constructed triangle’s angle between the same two sides — which was built to be exactly 90 degrees. The one genuinely new move is building that second triangle on purpose, specifically because it’s guaranteed to carry the angle being asked about, so SSS can hand it across to the original.

One to watch out for.

This is why the theorem and its converse needed two completely different proofs. The forward direction started from a right angle and found an area identity; the converse starts from the area identity and has to construct a right angle from scratch to borrow one.

Checking side lengths against the converse

A triangle has sides 5, 12 and 13. Is it a right triangle?

$5^2 + 12^2 = 25 + 144 = 169 = 13^2$ , so yes — the angle opposite the side of length 13 is a right angle.

A triangle has sides 6, 8 and 11. Is it a right triangle?

$6^2 + 8^2 = 36 + 64 = 100$ , but $11^2 = 121$ . The two do not match, so no angle in this triangle is a right angle.

This closes the area chapter: from counting squares in a rectangle, through triangles and parallelograms and trapezia, to the one relationship among the three that turns out to say everything about whether a triangle has a right angle at all.

Do the exercise

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