Exercise 45 Page 466

a We are asked to draw a figure of a right triangle that also shows the altitude to the hypotenuse.

We are also asked to formalize the conjecture that the product of the lengths of the two legs is equal to the product of the length of the hypotenuse and the length of the altitude to the hypotenuse.

Given : Prove : △ A B C is a right triangle with altitude to the hypotenuse \overline{C F} A C \cdot C B = A B \cdot C F

b To prove the conjecture of Part A, let's recall that accroding to Theorem

7 - 3

, the altitude to the hypotenuse divides a right triangle into two triangles that are similar to the original triangle. Let's use one of these similarities.

△ A B C \sim △ C B F

Notice that

△ C B F

has sides appearing in the conjecture from Part A.

The hypotenuse $\overline{C B}$ of $△ C B F$ is a leg of $△ A B C .$
The leg $\overline{C F}$ of $△ C B F$ is the altitude to the hypotenuse of $△ A B C .$

Let's identify the sides in $△ A B C$ that correspond to these sides of $△ C B F .$

Side in $△ C B F$	Corresponding side in $△ A B C$
$\overline{C B}$	$\overline{A B}$
$\overline{C F}$	$\overline{A C}$

We know that corresponding sides of similar triangles are proportional.

A C : C F = A B : C B

According to the Cross Products Property, the product of the extremes is equal to the product of the means in this proportion.

A C \cdot C B = A B \cdot C F

This completes the proof, so the conjecture is true. We can summarize the steps above in a flow proof.

Completed Proof

Given : Prove : △ A B C is a right triangle with altitude to the hypotenuse \overline{C F} A C \cdot C B = A B \cdot C F

Proof:

Subchapter links

Lesson Check p.464

Practice and Problem-Solving Exercises p.465-467

Standardized Test Prep p.467

Mixed Review p.467