What's the measure of the segment $PQ$ if the side of square $ABCD$ measures $a$?

Question

For refernce: In the figure, $P$ and $Q$ are points of tangency. Calculate $PQ$ if the side of square $ABCD$ measures $a$.

My progress:

$AQ = AD (tangents)\\\therefore AQ = AD = a\\similarly; AP = AQ = a\\ \therefore \triangle APD: \triangle AQG: \triangle APQ \rightarrow isosceles\\ EG =GH= \frac{a}{2}\\ \angle AGD = 45^o \therefore AG = \frac{a\sqrt2}{2}$

any hint where to use PQ...

Answer 1

Please note $EQ \perp AQ, ED = EQ, AD = AQ$. So, $\triangle DAE \sim \triangle QAE $ and $\angle DAE = \angle QAE = \alpha$ (say).

Then, $\angle PAI = 2 \alpha - 45^\circ$

$PQ = 2 PI = 2 AP \sin (2 \alpha - 45^\circ)$

$PQ = \sqrt2 ~a (\sin 2\alpha - \cos 2\alpha) \tag1$

Now note in $\triangle ADE$ that $\cos \alpha = \frac{2}{\sqrt5}, \sin \alpha = \frac{1}{\sqrt5}$

$\implies \sin 2\alpha = \frac{4}{5}, \cos 2\alpha = \frac{3}{5}$

Plugging into $(1)$, $PQ = \frac{a \sqrt2}{5}$

Answer 2

Alternatively, if you consider the area of the kite $ADEQ$ you get $$\overline{AD}\cdot \overline{DE} = \frac{\overline{AE}\cdot \overline{DQ}}2,$$ hence $\overline{DQ} = \frac2{\sqrt 5} a.$ Pythagorean Theorem on $\triangle DCQ$ yields $\overline{CQ} = \frac1{\sqrt 5}a$. Now, calling $\overline{QI} = x$, and applying again Pythagorean Theorem on $\triangle CQI$ and $\triangle AQI$ gives the equation $$\sqrt{\frac15a^2 - x^2} + \sqrt{a^2-x^2} = \sqrt 2 a.$$ Squaring both sides and reordering yields $$2a^2 + 5x^2 = 5\sqrt{\frac15a^4-\frac65 a^2x^2 +x^4}.$$ Squaring again finally takes us to $$50a^2x^2-1 = 0$$ that is $x = \frac1{5\sqrt 2} a$. Hence $\overline{PQ} = 2x = \frac{\sqrt 2}5a$.