 # File:Pythagoras similar triangles simplified.svg

## Summary

Description
English: Simplified version of similar triangles proof for Pythagoras' theorem.

In triangle ACB, angle ACB is the right angle. CH is a perpendicular on hypotenuse AB of triangle ACB.
In triangle AHC and triangle ACB, ∠AHC=∠ACB as each is a right angle. ∠HAC=∠CAB as they are common angles at vertex A. Thus triangle AHC is similar to triangle ACB by AA test.
Thus, $\frac{AC}{AB}=\frac{AH}{AC}$ $\therefore {AC}^{2} = {AB}\times {AH}$

In triangle BHC and triangle ACB, ∠BHC=∠ACB as each is a right angle. ∠HBC=∠CBA as they are common angles at vertex B. Thus triangle BHC is similar to triangle BCA by AA test.
Thus, $\frac{BC}{AB}=\frac{BH}{BC}$ $\therefore {BC}^{2} = {AB}\times {BH}$ $\therefore {AC}^{2} + {BC}^{2} = ({AB} \times {AH}) + ({AB} \times {BH})$ $\therefore {AC}^{2} + {BC}^{2} = {AB} \times ({AH} + {BH})$ $\therefore {AC}^{2} + {BC}^{2} = {AB} \times {AB}$ $\therefore {AC}^{2} + {BC}^{2} = {AB}^{2}$ which is the Pythagoras theorem.
Date 2012-05-15 10:08 (UTC)
Source This file was derived from:

Reference:

1. (2011) GEOMETRY Standard X, Secretary, Maharashtra State Board of Secondary and Higher Secondary Education, Pune-411 004, pp. 22, 23
Author
• derivative work: Gauravjuvekar
• derivative work: Gauravjuvekar This vector image was created with Inkscape. This is a retouched picture, which means that it has been digitally altered from its original version. Modifications: Simplified and removed lower-case notations and related elements per some comments at Wikipedia:Featured picture candidates/Pythagoras similar triangles proof. The original can be viewed here: Pythagoras_similar_triangles.svg. Modifications made by Gauravjuvekar.

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