Mathematical Olympiad questions: Plane Geometry-5

Known: Non-isosceles $\triangle{abc}$, $BD, ce$ respectively is $AC, ab$ edge on the high, $F $ is $BC the midpoint on the edge, $EF, df$ the midpoint is $M, n$, $I $ is $MN $ on a point, and meet $AI \para Llel bc$.

Verification: $IA = if$.

Analysis:

$BD, the intersection of ce$ is $\triangle{abc}$ of the heart $H $, and easy to know $A, E, H, d$ Four points round, that is, $IA $ is the tangent of $\odot{o}$.

On the other hand, by the $M, n$ are $\odot{o}$ two tangent $ (FD, FE) in points, according to the nature of the root axis can be proven.

Prove:

Easy to know, $BD, the intersection of ce$ is $\triangle{abc}$ of the heart $H $, and $A, E, H, d$ Four points round, set the circle for $\odot{o}$, and linked to the $AH, de$.

By $F $ is $BC $ midpoint and $BD \perp ac$, available:

$\ANGLE{BDF} = \angle{dbc} = 90^{\circ}-\ANGLE{ACB} = \angle{dah}\rightarrow fd$ is the tangent of the $\odot{o}$.

The same can be $FE $ is the tangent of $\odot{o}$.

and $\because m,n$ respectively is $FE, fd$ among points, $\therefore$ link $OF $ easy license $OF \perp mn$.

Therefore, $MN $ is $\odot{o}$ with the fixed-point $I $ root axis, which satisfies $IA ^2 = If^2\rightarrow IA = if$.

q$\cdot$ e$\cdot$ D

Commentary:

1. The root axis refers to the equal power of the two circles (trajectory), that is, a point to the power of the two circumference of the same points of the trajectory. The trajectory is a straight line perpendicular to the two-circle concentric line. In particular,

When the two circles intersect, the root axis is two round common chord;

When the two circles are tangent, the root axis is the Gongsche of two tangent points;

When the two circles are absent, the root axis is a straight line over the midpoint of the two round four Gongsche.

2. The radius of $\odot{f}$ in the subject is 0, so $I $ to $\odot{f}$ power for $IF ^2$.

Mathematical Olympiad questions: Plane Geometry-5