Square - Triangle - Geometry

Question

L is a point on the diagonal $AC$ of a square $ABCD$ such that $AL=3 \times LC$. $K$ is the midpoint of $AB$. What is the measure of angle $KLD$?

I tried solving the problem as :

$(AC)^2=(AB)^2)+((BC)^2$ -> by Pythagoras Theorem.

$ \Rightarrow(AL)^2)=(9/4)(AK)^2$ ---> Equation 1

Also, $(AC)^2=(AD)^2+(DC)^2$ -> by Pythagoras Theorem.

$ \Rightarrow(AL)^2=(9/8)(AD)^2$ ---> Equation 2

$\Rightarrow 2(AK)^2=(AD)^2$ From here, I am not able to solve the problem. ... Please advise.

Answer 1

Let $M$ be the feet of the altitude dropped from $L$ to $AB$. Then $AM=3MB$. Thus, $KM=MB$ and $\triangle KLB$ is isosceles. As $B$ is the reflection of $D$ wrt $AC$, $\angle DLA = \angle BLA$. Now it is easy to see that $$\angle KLD=\angle DLA+\angle ALK=\angle BLA+(\angle BKL-\angle KAL)=\angle BLA + \angle LBK-\angle BAL=(\angle BLA+\angle LBA)-\angle BAL=(180^{\circ}-\angle BAL)-\angle BAL=180^{\circ}-2 \cdot 45^{\circ}=90^{\circ}.$$

Answer 2

Hint: You can compute $KD$ $$KD=\frac{\sqrt{5}}{2}a$$ further

$$AL=\frac{3}{4}AC$$ Then we find $$KL$$ by

$$KL^2=\frac{9}{16}AC^2+\frac{a^2}{4}-2\cdot \frac{3}{4}AC\frac{a}{2}\cos(\frac{\pi}{4})$$ by the same procedure we get

$$DL^2=\frac{9}{16}(\sqrt{2}a)^2+a^2-2\cdot \frac{3}{4}AC\cdot a\cos(\frac{\pi}{4})$$ Then you can compute your searched angle.

and $$AC=\sqrt{2}a$$

Answer 3

Hint:   let $P,Q$ be the projections of $L$ onto $AB$ and $AD$ respectively. Then $LP=LQ$ ($\,=\frac{3}{4}AB$) and $KP=DQ$ ($\,=\frac{1}{4}AB$), so the right triangles $\triangle LKP$ and $\triangle LDQ$ are congruent. Given that $LP \perp LQ$ it follows that $LK \perp LD\,$.