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
