M. Ouadjaout
ENP,Dépt. SF, Lab. LRE
Algeria
Then let LQ meet PK at J
The common perpendicular bisector of LP and KQ ia a diameter of C
>From symmetry J lies on this diameter.
But triangles ALK, JLP and JKQ are similar
: J lies on AMND
: J is mid-point of MN. QED.
Regards, Peter Scales.
I have not understood this passage of your prouf :
"But triangles ALK, JLP and JKQ are similar; J lies on AMND"
What means the point D ?
Mohamed
Thank you.
Mohamed
Regards, Peter Scales.
I note by (D) a line in my original text, not a point.
Best regards.