Pythagoras Theorem [Proof and Simple Applications with Converse]

231. The Converse Check · When squares match, the angle is right

If BC² = AB² + AC², then ∠A is exactly 90°.

∠A = 90°∠A = 90°AB = 200AB = 200AC = 150AC = 150BC = 250BC = 250BC² = 62500BC² = 62500AB²+AC² = 62500AB²+AC² = 62500ABC
Converse of Pythagoras: if the square on one side equals the sum of the squares on the other two (BC² = AB² + AC²), the triangle is right-angled, with the right angle opposite that longest side — here ∠A = 90°.

Stuck? Ask Guru

Selina ICSE: Pythagoras Theorem [Proof and Simple Applications with Converse]

What this lesson covers

Try to break it

Drag A, B, or C until the side-length readouts satisfy AB² + AC² = BC². The moment that equation holds, ∠A becomes exactly 90°. Pull the vertices away from that condition and ∠A drifts off 90°. The converse of Pythagoras: the equation holds exactly when the angle is right.

How you build it

Make a triangle satisfying the Pythagoras rule.

  • Point tool: mark point A — the right-angle corner.
  • Point tool: mark point B — AB is one leg.
  • Segment tool: join A to B.
  • Perp tool: click A (on AB), then click outward. This builds the exact right angle; leg AC lies on this perpendicular.
  • Arc tool: centre A, drag out to your chosen leg length — the arc crosses the perpendicular at C, so AC is the second leg.
  • Point tool: mark C where the arc meets the perpendicular. Angle A is now exactly 90°.
  • Segment tool: join A to C — the second leg.
  • Segment tool: join B to C — the hypotenuse, opposite the right angle. Now BC² = AB² + AC².

The proof, step by step

Prove that if BC² = AB² + AC², then ∠A is a right angle.

  • Construct a right triangle PQR such that PQ = AB, PR = AC, and ∠QPR = 90°.
  • By the Pythagoras Theorem, QR² = PQ² + PR². Substituting the equal sides, QR² = AB² + AC².
  • We are given that BC² = AB² + AC². Therefore, BC² = QR², which implies BC = QR.
  • Now, ΔABC and ΔPQR have all three corresponding sides equal (AB=PQ, AC=PR, BC=QR). By SSS Congruence, ΔABC ≅ ΔPQR.
  • Since corresponding parts of congruent triangles are equal, ∠BAC = ∠QPR = 90°. Hence proved.

Worked example

In ΔABC, AB = 5 cm, AC = 12 cm, and BC = 13 cm. Is ΔABC a right-angled triangle? If yes, at which vertex?

Check: 5² + 12² = 25 + 144 = 169 = 13². Since AB² + AC² = BC², by the converse of Pythagoras theorem, ΔABC is right-angled at A.

  • Yes, at vertex A — correct
  • Yes, at vertex B
  • Yes, at vertex C
  • No, it is not right-angled
Hold to talk

Subscription Status