Pythagoras Theorem [Proof and Simple Applications with Converse]

230. The Similarity Shortcut · Pythagoras proved through matching triangles

AB² + BC² always equals AC², no matter where B sits on the arc.

Pythagoras can be proved using similar triangles: the altitude from the right angle splits the triangle into two smaller triangles, each similar to the original. Matching their proportional sides and adding the results gives AB² + BC² = AC².

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

What this lesson covers

Try to break it

Drag B around the arc. The triangle stretches into skinny or wide shapes, but the three sub-triangles (△ABC, △ABD, △ACD) stay similar to each other at every position. From that similarity, a² + b² = c² falls out automatically. Try to break the similarity — impossible.

How you build it

Inscribe a right triangle in a semicircle and drop the altitude to the hypotenuse.

Point tool: mark point A — one end of the hypotenuse.
Point tool: mark point C — the other end of the hypotenuse.
Segment tool: join A to C. AC is the hypotenuse and the diameter of the circle.
Midpoint tool: click A then C — O drops at the exact midpoint of AC.
Circle tool: centre O, open it out until it passes through A — the circle on AC as diameter.
Point tool: click anywhere on the circle's circumference — B snaps exactly onto the curve. Wherever B sits, angle ABC is exactly 90 degrees (angle in a semicircle).
Segment tool: join A to B.
Segment tool: join B to C.
Perp tool: click B, then click on the hypotenuse AC — the foot is D. Altitude BD splits triangle ABC into two triangles, each similar to the whole.

The proof, step by step

Prove that AB² + BC² = AC² in a right triangle.

In ΔABC and ΔBDC, ∠ABC = ∠BDC = 90° and ∠C is common. Therefore, ΔABC ~ ΔBDC by AA similarity.
From similarity, BC/AC = DC/BC, which gives BC² = AC × DC. (Equation I)
Similarly, in ΔABC and ΔADB, ∠ABC = ∠ADB = 90° and ∠A is common. Therefore, ΔABC ~ ΔADB by AA similarity.
From similarity, AB/AC = AD/AB, which gives AB² = AC × AD. (Equation II)
Adding Equation I and II: AB² + BC² = AC × AD + AC × DC = AC(AD + DC). Since AD + DC = AC, we get AB² + BC² = AC². Hence Proved.

Worked example

In right triangle PQR with ∠Q = 90°, PQ = 6 cm and QR = 8 cm. Find the length of PR.

By Pythagoras theorem, PR² = PQ² + QR² = 6² + 8² = 36 + 64 = 100. So PR = √100 = 10 cm.

10 cm — correct
11 cm
12 cm
14 cm