$ABCD$ is a square, with $K$ and $L$ are points on $BC$ and $DC$ respectively . If $AM \perp LK$ AND $\angle AKM = \angle AKB$ , Find $\angle LAK$ .

Question

Here is a diagram if needed :-

What I Tried :- I did angle-chasing , considered $\angle MKA = \angle AKB = x$ and then I did not get any information for some other angles , so I considered $\angle DAL$ to be $y$ . This didn't help , so overall this angle-chasing did not help . (Please don't ask me for my work in a diagram as it will take some time to send it. )

In the end, I saw that $\Delta AMK \cong \Delta ABK$ by $AAS$ congruency . This helps only a little bit as I only get to know that $AM = AB$ , nothing more .

Finally, I decided to use Geogebra to get some more information which I was missing for this Problem . There I found that $\Delta ADL \cong \Delta ALM \cong \Delta AMK \cong \Delta ABK$ , but I couldn't get it why it is so.

If I can show by some way that $\Delta ADL \cong \Delta ABK$ , then I can easily get $\angle LAK$ without any hesitation , but how will I do it?

Can anyone help?

Edit :- An Idea immediately struck my mind after I posted this problem (how unlucky).

I have that $\Delta AMK \cong \Delta ABK$ . This gives $AM = AB$ , and I was wrong to say that this is not a useful information .

From here I can show that $\Delta ADL \cong \Delta AML$ by $RHS$ congruency . ($AD = AM$ since $AB = AM$ and $AB = AD$ for $AB$ and $AD$ are the side-length of the square $ABCD$ , then we have the right-angle and $AL$ is common).

So I just moved $1$ step forward . Can I move another step ?

Answer 1

By the congruent triangles you have $\angle DAL=\angle LAM,\angle MAK=\angle KAB$.

These $4$ angles sum to $\angle DAB = 90^\circ$.

You are asked to find $\angle LAK = \angle LAM + \angle MAK$.