View Single Post
Old 2006-11-02, 16:01   #9
mfgoode
Bronze Medalist
 
mfgoode's Avatar
 
Jan 2004
Mumbai,India

205210 Posts
Lightbulb An easy one.


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
mfgoode is offline   Reply With Quote