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.

In square ABCD,

AB = BC = CD = DA

And, AK = BL = CM = DN

(All sides of a square are equal.) (Given)

So, AB - AK = BC - BL = CD - CM = DA - DN

⇒ KB = CL = DM = AN.......... (1)

In $△NAKand△KBL$
$\angle NAK=\angle KBL={90}^0$ (Each angle of a square is a right angle.)
AK=BL (Given)
AN=KB [From (1)]
So, by SAS congruence criteria,
\(\triangle NAK ≅ \triangle KBL \)
⇒NK=KL (Cpctc ...(2)
Similarly,
\(\triangle MDN≅ \triangle NAK \)

\(\triangle DNM≅ \triangle CML \)

\(\triangle MCL≅ \triangle LBK \)
\(\rightarrow MN = NK and \angle DNM=\angle KNA \) (Cpctc )… 3)
MN = JM and $\angle DNM=\angle CML$ (Cpctc )… 4)
ML = LK and $\angle CML=\angle BLK$ (Cpctc )… (5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML…….....(6)
And, $\angle DNM=\angle AKN=\angle KLB=\angle LMC$
Now,
In $△NAK$
$\angle NAK={90}^0$
Let $\angle AKN=x^0$

So, $\angle DNK={90}^0+x^0$ (Exterior angles equals sum of interior opposite angles.)
⇒$\angle DNM+\angle MNK={90}^0+x^0$

⇒$x^0+\angle MNK={90}^0+x^0$

⇒ $\angle MNK={90}^0$
Similarly,
$\angle NKL=\angle KLM=\angle LMN={90}^0$ ...(7)
Using (6) and (7), we get
All sides of quadrilateral KLMN are equal and all angles are \ ( 90^0 \)
So, KLMN is a square.