Given a rectangle $ABCD$ where $AB=|2a|$ and $BC= |\sqrt2a|$. On the side of $AB$, as a diameter, a semi-circle is constructed externally. Let $M$ be an arbitrary point on the semi-circle, the line $MD$ cuts $AB$ at $N$, and the line $MC$ cuts $AB$ at $L$. If $|AL|^2 + |BN|^2= \lambda a^2$, find $\lambda$.
Answer: $4$
My process: I started by taking a parametric point on the semi-circle and thereon started with my calculations, which were unnecessarily rigorous, can anyone suggest a more geometric approach to this problem?
Using coordinate geometry, from the given dimensions, we'll set
$ A = (-a, 0) $
$B = (a, 0) $
$C = ( a, \sqrt{2} a )$
$ D = (-a, \sqrt{2} a )$
Then point $M$ is given parametrically by
$ M = ( -a \cos t , - a \sin t )$
The parametric equation of line $MD$ is
$ p(t) = (- a, \sqrt{2} a ) + s ( - a \cos t + a , - a \sin t - \sqrt{2} a )$
To find the intersection with $AB$, set the $y$ coordinate to $0$. This gives
$ s = \dfrac{\sqrt{2} } { \sin t + \sqrt{2} } $
Therefore,
$ N = - a + \dfrac{ \sqrt{2} a (1 - \cos t) }{ \sin t + \sqrt{2}} $
Combining the two terms,
$ N = \dfrac{ - a ( \sin t + \sqrt{2} \cos t ) }{\sin t + \sqrt{2} } $
Likewise, the parametric equation of $MC$ is
$q(t) = (a , \sqrt{2} a) + s ( - a \cos t - a , - a \sin t - \sqrt{2} a ) $
At $y = 0$ , we get
$ s = \dfrac{\sqrt{2} } { \sin t + \sqrt{2} } $
Therefore,
$L = a - \dfrac{ \sqrt{2} a (1 + \cos t ) }{ \sin t + \sqrt{2} } $
Combining the two terms
$ L = \dfrac{ a ( \sin t - \sqrt{2} \cos t ) }{ \sin t + \sqrt{2} } $
Now,
$ A L = L - A = \dfrac{ a ( 2 \sin t - \sqrt{2} (\cos t - 1) ) }{\sin t + \sqrt{2}} $
$ BN = N - B = \dfrac{ -a ( 2 \sin t + \sqrt{2} (\cos t + 1 ) }{ \sin t + \sqrt{2} } $
Squaring both $AL$ and $BN$ and adding results in a straightforward fashion in,
$|AL|^2 + | BN |^2 = a^2 \left( \dfrac{ 8 \sin^2 t + 4 + 4 \cos^2 t + 8 \sqrt{2} \sin t }{ \sin^2 t + 2 + 2 \sqrt{2} \sin t } \right)$
This simplifies to,
$|AL|^2 + | BN |^2 = a^2 \left( \dfrac{ 4 \sin^2 t + 8 + 8 \sqrt{2} \sin t }{ \sin^2 t + 2 + 2 \sqrt{2} \sin t } \right)$
comparing the numerator and denominator, we finally get
$ | AL |^2 + |BN |^2 = 4 a^2 $
Thus $\lambda = 4 $