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 then prove that klmn is a square

Answer 1

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.

Answer 2

As ABCD is a square,
AB = BC = CD = DA     ...(1)
AK = BL = CM = DN   .....(2)
Subtracting (2) from (1)
AB- AK = BC - BL = CD - CM = DA - DN
BK = CL = DM = AN   ...(3)
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
Therefore KLMN is a square.