Question

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

Answer 1

​Given: In square ABCD, AK = BL = CM = DN.

To prove: KLMN is a square.

Proof:

In square ABCD,

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

$\Rightarrow$ KB = CL = DM = AN ...(1)

In $∆NAK$ and $∆KBL$,

$\angle NAK=\angle KBL=90°$ (Each angle of a square is a right angle.)

$AK=BL$ (Given)

$AN=KB$ [From (1)]

So, by SAS congruence criteria,

$∆NAK\cong ∆KBL$

$\Rightarrow NK=KL$ (CPCT) ...(2)

Similarly,

$$\begin{gathered} ∆MDN\cong ∆NAK \\ ∆DNM\cong CML \\ ∆MCL\cong LBK \end{gathered}$$

$\Rightarrow$MN = NK and $\angle DNM=\angle KNA$ (CPCT) ...(3)
MN = JM and $\angle DNM=\angle CML$ (CPCT) ...(4)
ML = LK and $\angle CML=\angle BLK$ (CPCT) ...(5)

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

NK = KL = MN = ML ...(6)

And, $\angle DNM=\angle AKN=\angle KLB=LMC$

Now,

In $∆NAK$,

$\angle NAK=90°$

Let $\angle AKN=x°$

So, $\angle DNK=90°+x°$ (Exterior angles equals sum of interior opposite angles.)

$$\begin{gathered} \Rightarrow \angle DNM+\angle MNK=90°+x° \\ \Rightarrow x°+\angle MNK=90°+x° \\ \Rightarrow \angle MNK=90° \end{gathered}$$

Similarly,
$\angle NKL=\angle KLM=\angle LMN=90°$ ...(7)

Using (6) and (7), we get

All sides of quadrikateral KLMN are equal and all angles are 90$°$.

So, KLMN is a square.