অতিভুজ-বাহু সাদৃশ্য

Suppose that we have two right triangles $\triangle ABC$ and $\triangle DEF$, with right angles $\angle C$ and $\angle F$. Also, suppose that we know that: $$\frac{AB}{DE}=\frac{AC}{DF}=k$$ for some positive real number $k$.

We would like to prove that $\triangle ABC\sim\triangle DEF$. There are a number of ways to do this; one of them is shown here.

The first way is to use the Pythagorean Theorem. We have the following two equations: $$AB^2=AC^2+BC^2$$ $$DE^2=DF^2+EF^2$$ Substitute $AB=k\cdot DE$ and $AC=k\cdot DF$ into the first equation: $$k^2 DE^2=k^2 DF^2+BC^2$$ Multiplying the second equation by $k^2$, $$k^2 DE^2=k^2 DF^2+k^2 EF^2$$ Using the two equations directly above, $$BC^2=k^2 EF^2$$ $$BC=k\cdot EF$$ Now, we know that: $$\frac{AB}{DE}=\frac{AC}{DF}=\frac{BC}{EF}=k$$ so we may use the SSS Similarity Theorem and we are done.

Since our choices of right triangles and legs were arbitrary, this will work with any two right triangles that have the hypotenuse and one of the legs in proportion. Thus, we have proved the HL Similarity Theorem.