Question

11. In two lines I and m are parallel and M is the
mid-point of the segment AB, where A is on l and B is on m.
Prove that any line segment through M with end points on
these parallel lines is bisected at M.
[Hint: Draw any line CD through M with Con land D on m.
Prove that A ACM = A BDM.)
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Answer 1

Answer:

Step-by-step explanation:

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