Equality implies equality

Myth

Let $D$ be a point on side $BC$ of triangle $ABC$ and point $P$ on segment $AD$ . A line through $D$ intersects segments $AB$ , $PB$ at $M$ , $E$ respectively, also intersects the extensions of segments $AC$ , $PC$ at $F$ , $N$ respectively.
If $DE = DF$ , prove that $DM =DN$ .

**Posted:** Wed Feb 09, 2005 9:51 am

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Myth is out of here

darij grinberg

Simple exercise in projective geometry.

Denote the line through the points D, M, E, F and N by d.

The Desargues involution theorem states that, if X, Y, Z, W are any four points and g is any line in the plane, then there exists an involution on the line g such that the pairs of points $\left(XY\cap g;\;ZW\cap g\right)$ , $\left(YZ\cap g;\;WX\cap g\right)$ and $\left(XZ\cap g;\;YW\cap g\right)$ are conjugate pairs in this involution. Applying this to the points X = A, Y = B, Z = C and W = P and the line g = d, we conclude that there exists an involution i on the line d such that the pairs of points $\left(AB\cap d;\;CP\cap d\right)$ , $\left(BC\cap d;\;PA\cap d\right)$ and $\left(AC\cap d;\;BP\cap d\right)$ are conjugate pairs in this involution i. But $AB\cap d=M$ , $CP\cap d=N$ , $BC\cap d=D$ , $PA\cap d=D$ , $AC\cap d=F$ and $BP\cap d=E$ . Hence, we obtain that the pairs of points (M; N), (D; D) and (F; E) are conjugate pairs in this involution i. In other words, the points M and N are conjugate to each other in the involution i, the point D is conjugate to itself in the involution i, and the points F and E are conjugate to each other in the involution i.

Of course, the result that the point D is conjugate to itself in the involution i simply means that the point D is a fixed point of the involution i. Thus, the involution i has a fixed point, i. e. it is a hyperbolic or parabolic involution. Let D' be the second fixed point of this involution i (in the degenerate case when the involution i is parabolic, i. e. has only one fixed point, set D' = D). It is well-known that any two conjugate points in an involution are harmonic conjugates with respect to the segment joining the two fixed points of the involution. Since the two fixed points of our involution i are D and D', and since the points M and N are conjugate to each other in the involution i, we thus can conclude that the points M and N divide the segment DD' harmonically. Hence, as a consequence, the points D and D' divide the segment MN harmonically. Similarly, since the points F and E are conjugate to each other in the involution i, the points D and D' divide the segment FE harmonically. In other words, the point D' is the harmonic conjugate of the point D with respect to the segment FE.

But since DE = DF, the point D is the midpoint of the segment FE, and thus, its harmonic conjugate D' with respect to the segment FE is an infinite point. But since the points D and D' divide the segment MN harmonically, the point D is the harmonic conjugate of the point D' with respect to the segment MN; since the point D' is an infinite point, its harmonic conjugate D must therefore be the midpoint of the segment MN. Thus, DM = DN. And the problem is solved.

Darij

**Posted:** Wed Feb 09, 2005 10:14 am

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Soarer

Applying menelaus theorem three times.
$\frac{{DP}}{{PA}}\frac{{AB}}{{BM}}\frac{{ME}}{{ED}} = 1$
$\frac{{AP}}{{PD}}\frac{{DN}}{{NF}}\frac{{FC}}{{CA}} = 1$
$\frac{{FC}}{{CA}}\frac{{AB}}{{BM}}\frac{{MD}}{{DF}} = 1$
Combining them we have
$\frac{{DN}}{{DM}} = \frac{{NF}}{{MD}}$
which gives the result.

**Posted:** Wed Feb 09, 2005 10:15 am

Myth

Great, siuhochung!

**Posted:** Wed Feb 09, 2005 10:17 am

_________________
Myth is out of here