euclidean-geometrygeometrytriangle-centrestriangles
I found out this problem when I was trying to solve another geometry problem but I stuck to solve it. Please help me. Thanks
Let $ABC$ be an acute triangle with $M$ be the midpoint of $BC$, $AK$ be the altitude from vertex $A$ and $H$ be the orthocenter. $I$ be the projection of $H$ on $AM$.
Prove that $MC^2 = MI. MA $.
$\newcommand{arc}[1]{\stackrel{\Large\frown}{#1}}$ The simplest way to prove $AX\perp YZ$ is using the angle between intersecting secants theorem which states:
If two lines intersect inside a circle, then the measure of the angle formed is one half the sum of the measure of the arcs intercepted by the angle and its vertical angle.
Thus $$\begin {align} \angle ADZ&=\frac12\big (\arc{AZ}+\arc{XY}\big)=\frac12\big (\arc{AZ}+\arc{XC}+\arc{CY}\big)\\ &=\frac14\big (\arc{AB}+\arc{BC}+\arc{CA}\big)=\frac\pi2. \end {align}$$
Let $x$ be the common $A$-bisector/$B$-median/$C$-altitude.
In its role as the $A$-bisector, $x$ satisfies (by Stewart's Theorem and the Angle Bisector Theorem) $$b^2\,\frac{ac}{b+c}+c^2\,\frac{ab}{b+c}=a\left(x^2+\frac{a^2bc}{(b+c)^2}\right)\;\to\;x^2(b+c)^2=bc(a+b+c)(-a+b+c) \tag{1}$$
As the $B$-median (again by Stewart, or Apollonius), $$c^2\;\frac{b}{2}+a^2\;\frac{b}{2} = b \left(x^2+\frac{b^2}{4}\right)\quad\to\quad 4x^2 = 2a^2-b^2+2c^2 \tag{2}$$
As the $C$-altitude (invoking Heron's formula), $$\frac12cx=|\triangle ABC|\quad\to\quad4c^2x^2=(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \tag{3}$$
The non-linear system $(1)$, $(2)$, $(3)$ turns out to be quartic. Throwing it into Mathematica for numerical solution, and discarding non-positive values, gives two options:
$$(a,b,c,x) = \left(1,1,1,\frac12\sqrt{3}\right)\qquad (a,b,c,x) = \left(1, 1.2225\ldots, 0.2381\ldots, 0.3933\ldots\right) \tag{$\star$}$$
(The values in my comment to the question correspond to taking $c=1$.) The first solution is, of course, the equilateral triangle; the second is obtuse:
So, if we restrict ourselves to acute triangles, the equilateral becomes the only solution. $\square$
Best Answer
The triangles $BAK$ and $HCK$ are similar because rotating $BAK$ by 90° gives a triangle which three sides are parallel to the sides of $HCK$. It follows that $\frac{BK}{AK} = \frac{HK}{CK}$, or \begin{equation}AK.KH = BK.CK \end{equation}
Now the triangles $A I H$ and $A K M$ are similar because they have two equal angles, hence $\frac{AI}{AH} = \frac{AK}{AM}$, hence \begin{align} AM.MI &= AM^2 - AM.AI\\ &= AM^2 - AK.AH\\ &= AM^2 - AK(AK-KH)\\ &= MK^2 + BK.CK\\ &=MK^2 + (MC-MK)(MC+MK)\\ &= MC^2 \end{align}