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### Parallel segments - not difficult

Inviato: 02 set 2009, 23:01
After having seen and loged in at your wonderfull site, I put a simple geometry problem for students of class 10-12..

Problem

The bisector of the internal angle A and the bisector of the external angle B of a triangle ABC meet the circumcircle of the triangle at points M,N.If I is the incenter of triangle ABC , prove that MN is parallel to CI .

Babis

Inviato: 03 set 2009, 17:53
Nice problem stergiu

Let $\alpha,\beta,\gamma$ be the angles $\angle CAB,\angle ABC,\angle BCA$.

First case: M belongs to the arc AB wich not contains C.

$\angle MNA = \angle MBA$ because they lie on the same arc.
Since $M$ is on the bisector of the external angle B, $\angle MBA = \dfrac{(\angle CAB + \angle BCA)}{2} = \dfrac{\alpha + \gamma}{2}$.

So $\angle MNA = \dfrac{\alpha + \gamma}{2}$.

In the triangle $AIC$, since $\angle CIN$ is the external angle of $\angle AIC$ , it equals $\angle IAC + \angle ACI = \dfrac{\alpha + \gamma}{2}$.

From $\angle CIN = \angle MNA$ we conclude that CI is parallel to MN,

Second case: M belongs to the arc BC wich not contains A

The bisector of the external angle $B$ meets the circumcircle in a point $M$ such that $AM=MC$(if you draw the bisector of the internal angle $B$ you'll notice that this bisector meets the circumcircle at a point T such that AT=TC. Since BM is perpendicular to BT, M divides the arc AC in two equal parts). So $\angle MNA = \angle MBC = \dfrac{\alpha + \gamma}{2}$.

From $\angle CIN = \angle MNA = \dfrac{\alpha + \gamma}{2}$ we conclude that CI is parallel to MN.

Inviato: 03 set 2009, 18:15
The solution is ok, but I suppose that you meant $\dfrac{\alpha+\gamma}{2}$ in all the passages, instead you put $\beta$...

Inviato: 03 set 2009, 18:57
Yes, you're right. I corrected it

Inviato: 04 set 2009, 10:12
GioacchinoA ha scritto:Nice problem stergiu

Let $\alpha,\beta,\gamma$ be the angles $\angle CAB,\angle ABC,\angle BCA$.

First case: M belongs to the arc AB wich not contains C.

$\angle MNA = \angle MBA$ because they lie on the same arc.
Since $M$ is on the bisector of the external angle B, $\angle MBA = \dfrac{(\angle CAB + \angle BCA)}{2} = \dfrac{\alpha + \gamma}{2}$.

So $\angle MNA = \dfrac{\alpha + \gamma}{2}$.

In the triangle $AIC$, since $\angle CIN$ is the external angle of $\angle AIC$ , it equals $\angle IAC + \angle ACI = \dfrac{\alpha + \gamma}{2}$.

From $\angle CIN = \angle MNA$ we conclude that CI is parallel to MN,

Second case: M belongs to the arc BC wich not contains A

The bisector of the external angle $B$ meets the circumcircle in a point $M$ such that $AM=MC$(if you draw the bisector of the internal angle $B$ you'll notice that this bisector meets the circumcircle at a point T such that AT=TC. Since BM is perpendicular to BT, M divides the arc AC in two equal parts). So $\angle MNA = \angle MBC = \dfrac{\alpha + \gamma}{2}$.

From $\angle CIN = \angle MNA = \dfrac{\alpha + \gamma}{2}$ we conclude that CI is parallel to MN.
Nice solution !

Let me only give a hint for another solution :

If the bisectors of external angles A,B,C define the triangle KLR( K is the ex- center for angle C and L the ex- center for angle B), then I is obviously the orthocenter of triangle KLR and the circumcircle (C) of triangle ABC is the Euler(nine point) circle for triangle KLR.

This means that this circle (C) passes trhough the midpoints of KL,LR ,RK. Hence M is the midpoint of KR .But (C) passes also from the midpoint N of IR.Thus in triangle IKR segment MN is parallel to IK ect.

I must say that my first solution to this(own ) problem is like Gioacchino A's, but I give this hint to point out these nice results for the nine point circle and the triangle of ex -centers.

Have a nice day. I hope I will send you later one more nice problem.Remember that geometry is the best branch in school maths(in olympiads too!)

Babis

Inviato: 04 set 2009, 10:54
stergiu ha scritto:Remember that geometry is the best branch in school maths(in olympiads too!)
Omg..