AIR MATH

Math Question

The altitude of a right triangle divides the triangle into two triangles similar to each other and to the original triangle. So, in the diagram below, \( \triangle P Q R \sim \triangle P S Q \sim \triangle Q S R \). Complete the proof that \( P Q^{2}+Q R^{2}=P R^{2} \). Since \( \triangle P Q R \sim \triangle Q S R \), then \( \frac{P R}{Q R}=\quad \nabla \).You can rewrite this equation as \( \nabla \). Similarly, since \( \triangle P Q R \sim \triangle P S Q \), then \( \frac{P Q}{P R}= \) You can rewrite this equation as Now, using, the Addition Property of Equality, \( P Q^{2}+Q R^{2}= \) Then, by substitution, \( P Q^{2}+Q R^{2}= \) -. So, \( P Q^{2}+Q R^{2}= \) by the Distributive Property. Because \( P S+S R=P R \) by the Additive Property of Length, using substitution. Therefore, \( P Q^{2}+Q R^{2}=P R^{2} \).

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