Distance ratio on a triangle side for angle bisecting line, altitude and contact point of the inscribed circle

Question

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How can I proof $$|MD|\times |MF| = |ME|^2$$

I thought about the Geometric mean theorem ($h^2=pq$). But how to draw the matching triangle?

Answer 1

Denote by $a, b, c$ the lengths of $BC, AC, AB$.

$AD = c\cdot \frac{b}{a+b} $ by the angle bisector theorem. So, $$MD= |\frac{bc}{a+b} - \frac{c}{2} |= c (|\frac{b}{a+b} - \frac{1}{2}|)=c\frac{|b-a|}{2(a+b)}$$

$$MF = |b\cos \alpha -\frac{c}{2}| = |b \frac{b^2+c^2-a^2}{2bc} - \frac{c}{2}|= \frac{|b^2-a^2|}{2c}$$

$AE=\frac{b+c-a}{2}$, so $$ME=|\frac{b+c-a}{2}-\frac{c}{2}|= \frac{|b-a|}{2}$$

Now, multiply the first two segments

$$MD \cdot MF = c \cdot \frac{|b-a|}{2(a+b)} \cdot \frac{|b^2-a^2|}{2c} = (\frac{b-a}{2})^2 = ME^2 $$