Perpendicular in Right-Angled Triangle makes two Similar Triangles

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Theorem

In the words of Euclid:

If in a right-angled triangle a perpendicular be drawn from the right angle to the base, the triangles adjoining the perpendicular are similar both to the whole and to one another.

(The Elements: Book $\text{VI}$: Proposition $8$)


Proof

Let $\triangle ABC$ be a right-angled triangle such that $\angle BAC$ is a right angle.

Let $AD$ be drawn perpendicular to $BC$.

We need to show that $\triangle ABD$ is similar to $\triangle CAD$ which is similar to $\triangle CBA$ which is in turn similar to $\triangle ABD$.

Euclid-VI-8.png

We have that $\angle BAC = \angle ADB$ as both are right angles.

Then $\angle ABC$ is common to both $\triangle ABC$ and $\triangle ABD$.

So from Sum of Angles of Triangle Equals Two Right Angles:

$\angle ACB = \angle BAD$

So $\triangle ABC$ is equiangular with $\triangle ABD$.

From Equiangular Triangles are Similar it follows that $\triangle ABC$ is similar to $\triangle ABD$.

Similarly we see that $\triangle ABC$ is equiangular with $\triangle ACD$.

Therefore each of $\triangle ABD$ and $\triangle ACD$ are similar to $\triangle ABC$.

Finally, $\triangle ABD$ and $\triangle ACD$ are equiangular with each other and so similar to each other.

$\blacksquare$


Porism

In the words of Euclid:

From this it is clear that, if in a right-angled triangle a perpendicular be drawn from the right angle to the base, the straight line so drawn is a mean proportional between the segments of the base.

(The Elements: Book $\text{VI}$: Proposition $8$ : Porism)


Historical Note

This proof is Proposition $8$ of Book $\text{VI}$ of Euclid's The Elements.


Sources

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