Euclid's first theorem, proofs and converse

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Consider Euclid’s first theorem:

• (a) state the theorem;
• (b) prove it using equivalence;
• (c) prove it using similarity;
• (d) prove the converse: if a triangle satisfies the property of Euclid’s first theorem, then it is right-angled (use the similarity criteria).

Solution

Let $ABC$ be right-angled at $C$, with $CH$ the altitude to the hypotenuse $AB$; $AH$ and $BH$ are the projections of the legs $AC$ and $BC$.

(a) Statement. In a right triangle, the square on a leg is equivalent to the rectangle whose sides are the hypotenuse and the projection of that leg onto it. In similarity form, each leg is the geometric mean between the hypotenuse and its own projection: $\overline{AC}^{\,2}=\overline{AB}\cdot\overline{AH},\qquad \overline{BC}^{\,2}=\overline{AB}\cdot\overline{BH}.$

(b) Proof by equivalence. One builds the square on $AC$ and the rectangle with sides $AB$ and $AH$, and shows, via suitable translations/decompositions (parallelograms with the same base between the same parallels are equivalent), that both are equivalent to the same parallelogram; hence $\overline{AC}^{\,2}=\overline{AB}\cdot\overline{AH}.$

(c) Proof by similarity. Triangles $ACH$ and $ABC$ share angle $\hat A$ and both have a right angle ($\widehat{AHC}=\widehat{ACB}=90^\circ$): by the AA criterion $ACH\sim ABC$. From the proportion of corresponding sides $\frac{\overline{AC}}{\overline{AB}}=\frac{\overline{AH}}{\overline{AC}}\ \Longrightarrow\ \overline{AC}^{\,2}=\overline{AB}\cdot\overline{AH}.$

(d) Converse. Hypothesis: in $ABC$ there is $H\in AB$ with $CH\perp AB$ and $\overline{AC}^{\,2}=\overline{AB}\cdot\overline{AH}$. Then $\dfrac{\overline{AC}}{\overline{AB}}=\dfrac{\overline{AH}}{\overline{AC}}$, and triangles $ACH$ and $ABC$ share angle $\hat A$ between proportional sides: by the SAS similarity criterion $ACH\sim ABC$. Hence $\widehat{ACB}\cong\widehat{AHC}=90^\circ$: triangle $ABC$ is right-angled at $C$. $\;\blacksquare$ $\boxed{\overline{AC}^{\,2}=\overline{AB}\cdot\overline{AH}\iff \widehat{ACB}=90^\circ.}$