Well fetofs I have come up with a very elegant derivation of angle MNR

Which is required.

In triangle ABC drop a perpendicular from C to AB and call it P.

Drop a perpendicular from A to BC intersecting CP at H (the orthocentre)

In triangle PBC, angle PBC = 70* (given)

Ang. BPC = 90* (construction)

Therefore Ang. BCP = 20*

Now line MN ( M and N being midpoints) is parallel (//) to base BC in Triangle ABC.

In triangle AHC , R and N are midpoints.

Therefore RH is // to HC

Therefore Ang. MNR = Ang. BCP = 20*

Because of being angle between //’s.

Q.E.D.

Mally