65. Parallel to BC through I .
by Virgil Nicula, Jul 22, 2010, 1:12 AM
Nice parallel to BC through I (<== click).
Lemma (new). Points
(in this order) is harmonical division on line 
,
where
is midpoint of ![$[CD]$](//latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
.
Proof.
. Observe that 
and 
. Prove analogously and second equivalence.
Proof. Prove easily that 
is an harmonical division . Indeed, 
. Since is midpoint of 
obtain 
. Apply Menelaus's theorem to transversal 
for 
. From relations 
and 
obtain that 
is a parallelogram.
Lemma (new). Points
(in this order) is harmonical division on line 
,where
is midpoint of ![$[CD]$](http://latex.artofproblemsolving.com/e/7/0/e70960e9e5738a46ad23f794e796ef3cb4ad7e2c.png)
.Proof.
. Observe that 
and 
. Prove analogously and second equivalence.Sunken Rock wrote:
Let 
be a triangle with incenter 
and circumcircle 
. Denote 
and 
. For 
define midpoint 
of ![$[IM]$](//latex.artofproblemsolving.com/5/2/0/520a4d2e5d1bc639a6b51d0b8ec17fb55b9949e6.png)
, 
and 
. Prove that is a parallelogram.
be a triangle with incenter 
and circumcircle 
. Denote 
and 
. For 
define midpoint 
of ![$[IM]$](http://latex.artofproblemsolving.com/5/2/0/520a4d2e5d1bc639a6b51d0b8ec17fb55b9949e6.png)
, 
and 
. Prove that 
is a parallelogram.
is an harmonical division . Indeed, 
. Since is midpoint of 
obtain 
. Apply Menelaus's theorem to transversal 
for 
. From relations 
and 
obtain that is a parallelogram.This post has been edited 14 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:08 PM
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