# ABCD is a square. P and Q are points on BC and CD, respectively, such that BP=3PC and CQ=QD. Prove that AQ=2PQ and APQ=CPQ?

Mar 28, 2018

Use the Pythagorean Theorem to show that $A Q = 2 P Q$ and DeltaAPQ~DeltaCPQ

#### Explanation: Assume that the length of a side of square ABCD is 4 units. Then

$B P = 3$
$P C = 1$
$Q C = 2$
$D Q = 2$
$A D = 4$
$A B = 4$

Consider the right triangle $P Q C$ where angle $C$ is 90 degrees. $P C$ and $Q C$ are the legs of the right triangle. Since $P C = 1$ and $Q C = 2$, by the Pythagorean Theorem, the length of the hypotenuse $P Q = \sqrt{{1}^{2} + {2}^{2}} = \sqrt{5}$.

Now consider the right triangle $A D Q$ where angle $D$ is 90 degrees. $A D$ and $D Q$ are the legs of this triangle. Since $A D = 4$ and $D Q = 2$, by the Pythagorean Theorem, the length of the hypotenuse $A Q = \sqrt{{4}^{2} + {2}^{2}} = \sqrt{20}$.

So

$\frac{A Q}{P Q} = \frac{\sqrt{20}}{\sqrt{5}} = \sqrt{4} = 2$.

Therefore

$A Q = 2 P Q$.

Now consider right triangle $A B P$ where angle $B$ is 90 degrees. $A B$ and $B P$ are the legs of of the right triangle. Since $A B = 4$ and $B P = 3$, the length of the hypotenuse $A P = \sqrt{{4}^{2} + {3}^{2}} = 5$.

Now consider triangle $A P Q$. Two of its sides are of length $\sqrt{5}$ and $\sqrt{20}$. Note that if we add squares the length of both of these sides, we get the square of length of the third side

${\left(\sqrt{5}\right)}^{2} + {\left(\sqrt{20}\right)}^{2} = {5}^{2}$

which means that triangle $A P Q$ must be a right triangle. We have already proven that the ratio of the length of its longer leg to its shorter leg is 2. The same is true for right triangle $C P Q$, so $A P Q$ and $C P Q$ must be similar (by the S ide- A ngle- S ide Theorem).