Proof of Euclid's theorems

The altitude $CH$ relative to the hypotenuse divides the right-angled triangle $ABC$ into two smaller triangles. We show that these are similar to each other: it is from this similarity that Euclid’s theorems follow.

The altitude $CH$ divides the right-angled triangle $ABC$ into the two triangles $ACH$ and $BHC$.

Proof — Euclid's second theorem

We compare the triangles $ACH$ and $BHC$. They have:

• $\widehat{AHC} \cong \widehat{BHC} = 90^\circ$ (right angles);
• $\widehat{ACH} \cong \widehat{BCH}$ (complementary angles of the same angle).

By the first similarity criterion (two congruent angles), the two triangles are similar.

From the similarity, the corresponding sides are in proportion: $\frac{AH}{CH} = \frac{CH}{BH}.$ ∎

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Topics: Similarity
Concepts: Similarity criteria · Euclid’s theorems · Similar triangles · Right-angled triangle
Skills: Proving · Synthetic geometry
People: Euclid