Gems of Geometry

In the figure above,

Let $AL = 3x$ and $LC = x$ .

Then by Pythagoras theorem in $\triangle ABC$

$\large\ { \Rightarrow { \left( 4x \right) }^{ 2 } = { \left( AB \right) }^{ 2 } + { \left( BC \right) }^{ 2 } \\ \Rightarrow { \left( AB \right) }^{ 2 } = 8{ x }^{ 2 } \\ \Rightarrow AB = 2\sqrt { 2 } x}$ .

So, that $AK = KB = x \sqrt 2$ . Also $AD = 2 \sqrt 2 x$ as $ABCD$ is a square.

As $AC$ is a diagonal, $\angle DAC = 45^0$ .

By cosine rule in $\triangle ALD$ ,

$\large\ {\Rightarrow \cos { 45 } = \frac { { \left( AD \right) }^{ 2 } + { \left( AL \right) }^{ 2 } - { \left( DL \right) }^{ 2 } }{ 2\left( AD \right) \left( AL \right) } \\ \Rightarrow \frac { 1 }{ \sqrt { 2 } } = \frac { 8x^{ 2 } + 9{ x }^{ 2 } - { \left( DL \right) }^{ 2 } }{ 12\sqrt { 2 } x } \\ \Rightarrow DL= x\sqrt { 5 }}$ .

Again applying Cosine rule in $\triangle AKL$ with $\angle KAL = 45^0$ , We get

$\large\ LK=x \sqrt 5$ .

Now by pythagoras theorem in $\triangle ADK$ , we have

$\large\ DK = x \sqrt { 10 }$ .

Observe in $\triangle LDK$

$\large\ { DK }^{ 2 } = 10{ x }^{ 2 } = 5{ x }^{ 2 } + 5{ x }^{ 2 } = { \left( LD \right) }^{ 2 } + { \left( LK \right) }^{ 2 }$ .

Which implies $\angle DLK = 90^0$ .

Also, as $LK = LD$

$\Rightarrow \angle KDL = 45^0$ .

$\angle DLK = 90^{o}$ so $DLKA$ cyclic and $\angle LDK = \angle LAK = 45^{o}$

We see that triangle DLA and MLC are similar. .'. CM = x/3. Now we copy the triangle DCM as DAM' . Then we find DM' and DK. Finally draw a perpependicular on DK to find angle M'DK. or just using cosine rule we can find that. .'. Our requIred angle is 45 degrees.