K L M and N are points on side AB BC CD and DA respectively of a squar

1.	K L M and N are points on side AB BC CD and DA respectively of a square ABCD such that AK=CM=DN. prove that KLMN is square
Answer» As ABCD is a square, so AB = BC = CD = DA ...(1) Also AK = BL = CM = DN .....(2) Subtracting (2) from (1) We get AB- AK = BC - BL = CD - CM = DA - DN BK = CL = DM = AN ...(3) So now we have AK = BL = CM = DN AN = BK= CL = DM Squaring and adding AK² + AN² = BL² + BK²= CM² + CL² = DN² + DM² ...(4) But <A = <B = <C = <D = 90° By Pythagorean theorem (4) Becomes KN² = KL² = LM² = NM² So KN = KL = LM = NM So KLMN is a rhombus But <1 = <3 as triangles are congruent And < 1 + <2 = 90 So <2 + <3 = 90 Hence <KNM = 90° Therefore KLMN is a square

K L M and N are points on side AB BC CD and DA respectively of a square ABCD such that AK=CM=DN. prove that KLMN is square

Answer»

As ABCD is a square, so

AB = BC = CD = DA ...(1)

Also

AK = BL = CM = DN .....(2)

Subtracting (2) from (1)

We get

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

BK = CL = DM = AN ...(3)

So now we have

AK = BL = CM = DN

AN = BK= CL = DM

Squaring and adding

AK² + AN² = BL² + BK²= CM² + CL² = DN² + DM² ...(4)

But <A = <B = <C = <D = 90°

By Pythagorean theorem (4) Becomes

KN² = KL² = LM² = NM²

KN = KL = LM = NM

So KLMN is a rhombus

But <1 = <3 as triangles are congruent

And < 1 + <2 = 90

So <2 + <3 = 90

Hence <KNM = 90°

Therefore KLMN is a square