Question

If triangle ABC, M and N are points on the sides AB and AC respectively such that BM = CN. If angle B= angle C, then show that MN||BC.

Answer 1

this question is the proof of midpoint theorem
and can be solved easily


To Prove: DE || BC and DE = 1212 BC.

Construction: Extend line segment DE to F such that DE = EF.

Proof: In △△ ADE and △△ CFE

AE = EC   (given)

∠∠AED = ∠∠CEF (vertically opposite angles)

DE = EF   (construction)

hence

△△ ADE ≅≅ △△ CFE (by SAS)

Therefore,
∠ADE = ∠CFE   (by c.p.c.t.)

∠DAE = ∠FCE   (by c.p.c.t.)

and AD = CF  (by c.p.c.t.)

The angles ∠ADE and ∠CFE are alternate interior angles assuming AB and CF are two lines intersected by transversal DF.

Similarly, ∠DAE and ∠FCE are alternate interior angles assuming AB and CF are two lines intersected by transversal AC.

Therefore, AB ∥ CF

So, BD ∥ CF

and BD = CF (since AD = BD and it is proved above that AD = CF)

Thus, BDFC is a parallelogram.

By the properties of parallelogram, we have

DF ∥ BC

and DF = BC

DE ∥ BC

and DE = 1212BC  (DE = EF by construction)

Hence proved.





I AM VERY VERY SORRY CAUSE IS COPIED THE WHOLE ANSWER AND CHANGED THE NAMING BUT WRITING THIS IS GOING TO TAKE A LOT TIME AND NOW I AM BUSY

SORRY AGAIN