Question

in the given figure L || M and M is the midpoint of a line segment AB show that M is also the midpoint of any line segment CD having its end point C and D on L and m respectively

Answer 1

In triangles ACM and BDM,
Angle MAC = Angle MBD(Alternate int. angles)
Angle ACM = Angle BDM(alternate int. angles)
Angle AMC = Angle DMB(vertically op. angles)

Thus,
Triangle ACM is congruent to Triangle BDM(AAA criterion)

Therefore, AM= MB(cpct)
CM=MD(cpct)

Therefore, M bisects CB
Hence proved

Answer 2

Answer:

Step-by-step explanation

Proof: In triangle AMC and triangle DMB

∠AMC =  ∠DMB (Vertically Opp. angles)

∠CAM = ∠MBD (Alternate interior angles)

Therefore, ΔAMC ≅ ΔDMB (By ASA rule)

∴ DM = MC (CPCT)

∴ M is the midpoint of DC

(HENCE PROVED)