K, L, M and N are points on the sides AB, BC, CD and DA respectively of a square

K, L, M and N are points on the sides ABBCCD and DA respectively of a square ABCD such that AK BL CM DN. Prove that KLMN is a square.



Given: In square ABCD, AK BL CM DN

To prove: KLMN is a square.

In square 

AB = BC = CD = DA             (All sides of a square are equal.)

And, AK BL CM DN     (Given)

So, AB  AK = BC  BL = CD  CM = DA  DN

 KB = CL = DM = AN        …(1)

In $\Delta N A K$ and $\Delta K B L$,

$\angle N A K=\angle K B L=90^{\circ}$ (Each angle of a square is a right angle.)

$A K=B L$                          (Given)

$A N=K B$                       [From (1)]


So, by SAS congruence criteria,

$\Delta N A K \cong \Delta K B L$

$\Rightarrow N K=K L \quad(\mathrm{CPCT}) \quad \ldots(2)$


$\Delta M D N \cong \Delta N A K$

$\Delta D N M \cong C M L$


$\Delta M C L \cong L B K$

$\Rightarrow M N=N K$ and $\angle D N M=\angle K N A \quad$ (CPCT) $\quad$…(3)

$M N=J M$ and $\angle D N M=\angle C M L \quad$ (CPCT) $\quad \ldots$ (4)

$M L=L K$ and $\angle C M L=\angle B L K \quad$ (CPCT) $\quad \ldots$ (5)

From (2), (3), (4) and (5), we get

NK = KL = MN = ML      …(6)

And, $\angle D N M=\angle A K N=\angle K L B=L M C$


In $\Delta N A K$,

$\angle N A K=90^{\circ}$

Let $\angle A K N=x^{\circ}$

So, $\angle D N K=90^{\circ}+x^{\circ}$ (Exterior angles equals sum of interior opposite angles.)

$\Rightarrow \angle D N M+\angle M N K=90^{\circ}+x^{\circ}$

$\Rightarrow x^{\circ}+\angle M N K=90^{\circ}+x^{\circ}$

$\Rightarrow \angle M N K=90^{\circ}$


$\angle N K L=\angle K L M=\angle L M N=90^{\circ}$    (7)

Using (6) and (7), we get
All sides of quadrikateral KLMN are equal and all angles are 90°°.

So, KLMN is a square.



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