Pythagoras Theorem and its converse

What is the Pythagoras Theorem? The Pythagoras Theorem applies to right-angled triangles (one angle exactly $90^\circ$). It states: in a right-angled triangle, the square of the hypotenuse (the side opposite the right angle, which is also the longest side) equals the sum of the squares of the other two sides.

Equation. $$a^2+b^2=c^2,$$ where $c$ is the hypotenuse and $a,b$ are the legs (base and height).

Helper result (perpendicular from the right angle). If a perpendicular is drawn from the right-angle vertex of a right triangle to the hypotenuse, then the two triangles on each side of the perpendicular are similar to the whole triangle and to each other. From this, three relations follow for $\triangle ABC$ right-angled at $B$ with foot $D$ on hypotenuse $AC$: $BD^2=AD\cdot DC$, $AB^2=AC\cdot AD$ and $BC^2=AC\cdot DC$.

Proof of Pythagoras Theorem (using similarity). Let $\triangle ABC$ be right-angled at $B$, and let $BD\perp AC$. Then $\triangle ADB\sim\triangle ABC$, giving $\dfrac{AD}{AB}=\dfrac{AB}{AC}$, so $AB^2=AC\cdot AD$. Also $\triangle BDC\sim\triangle ABC$, giving $\dfrac{CD}{BC}=\dfrac{BC}{AC}$, so $BC^2=AC\cdot CD$. Adding: $AB^2+BC^2=AC(AD+CD)=AC\cdot AC=AC^2$. Hence $AB^2+BC^2=AC^2$.

What is the Converse? If in a triangle the square of one side equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. So if the three sides satisfy $a^2+b^2=c^2$, the triangle is right-angled, with the right angle opposite the longest side $c$. This converse lets us decide whether a triangle is right-angled from its side lengths alone.

Proof of the Converse. Suppose in $\triangle ABC$ we have $AB^2+BC^2=AC^2$. Construct a separate right triangle $PQR$ right-angled at $Q$ with $PQ=AB$ and $QR=BC$. By the Pythagoras Theorem, $PR^2=PQ^2+QR^2=AB^2+BC^2=AC^2$, so $PR=AC$. Now $\triangle ABC$ and $\triangle PQR$ have all three pairs of sides equal, hence $\triangle ABC\cong\triangle PQR$ (SSS). Therefore $\angle B=\angle Q=90^\circ$, proving $\triangle ABC$ is right-angled at $B$.

Pythagorean triplets. Whole-number side sets that satisfy the theorem are called Pythagorean triplets: $(3,4,5)$, $(5,12,13)$, $(8,15,17)$, $(7,24,25)$, and their multiples such as $(6,8,10)$ and $(9,12,15)$. For any natural numbers $m>n$, the triple $\big(m^2-n^2,\ 2mn,\ m^2+n^2\big)$ is always Pythagorean. Recognising these saves a lot of calculation in the exam.

Standard structural results. (i) The diagonal of a square of side $a$ is $a\sqrt2$. (ii) The altitude of an equilateral triangle of side $a$ is $\dfrac{\sqrt3}{2}\,a$, so for side $2a$ the altitude is $\sqrt3\,a$. (iii) In an isosceles right triangle, the hypotenuse $=\sqrt2\times(\text{leg})$. (iv) For an equilateral triangle of side $a$, the relation $3\times(\text{side})^2=4\times(\text{altitude})^2$ often appears in proofs.

Advanced multi-triangle results. Several NCERT problems combine the theorem with a point inside or on a triangle. For instance, if $O$ is any point inside a rectangle $ABCD$, then $OA^2+OC^2=OB^2+OD^2$. In an isosceles right triangle right-angled at $C$ with $AC=BC$, the hypotenuse satisfies $AB^2=2\,AC^2$. When a perpendicular falls inside a right triangle, you can chain the three relations $BD^2=AD\cdot DC$, $AB^2=AC\cdot AD$, $BC^2=AC\cdot DC$ to find any missing segment. Mastering these patterns lets you tackle the harder board and Olympiad questions where the theorem is applied two or three times in one diagram.

Common mistakes. (i) Treating the longest side as a leg — always identify the hypotenuse first. (ii) Adding the sides instead of their squares. (iii) Forgetting to take the square root at the end. (iv) Applying the theorem to a triangle that is not right-angled. (v) Using the converse without first squaring the longest side.

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