Given: △akl, m∠k=90° ak = kl = 2 m∈ al , am = ml find: ml and km

All the trianges involved are isosceles right triangles. You know that the hypotenuse is the leg multplied by √2.

ML = KM = KL/√2 = 2/√2

ML = KM = √2

Remark

AKL is an isosceles right triangle. That's because AK = KL = 2 and K is a right angle.

That makes <A = <L = 45o

Since AM = ML (Given) and KM is common to both Triangle AKM and Triangle LKM, the two triangles are congruent by SSS

Since <AKM = <LKM = 90/2 = 45 (Corresponding parts of congruent triangles are congruent) KM is a perpendicular bisector of AL.

So KM = ML because <MKL and <KLM are both = 45 degrees.

If you find one of KM or ML, you have the other.

KM^2 + LM^2 = 2^2       Remember that KM is perpendicular to AL

2KM^2 = 4

KM^2 = 2

sqrt(KM^2) = sqrt(2)

KM = sqrt(2)