Segment ratio

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#1 h Sep 11, 2024, 4:21 AM

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Let $B$ and $C$ be points on a circle with center $A$ .
Let $D$ be a point on segment $AB$ .
Let $F$ be one of the intersections of the circle with center $D$ and passing through $B$ and the circle with diameter $DC$ .
Prove that $\dfrac {AD}{AC} = \dfrac {CF^2}{CB^2}$ .

This post has been edited 1 time. Last edited by xeroxia, Sep 11, 2024, 4:21 AM

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#2 h Sep 11, 2024, 4:03 PM

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Please view the attachment. Let $CF$ intersects $\odot A$ at another point $C_2$ , and $BF$ intersects $\odot A$ at another point $S$ .
From the condition we know that $CF$ is tangent to $\odot D$ , hence $S$ should be the midpoint of $\stackrel\frown{CC_2}$ (Because $\odot A$ and $\odot D$ are homothetic).
From $DF \perp CF$ , $AS\perp CF$ we have $AS\parallel DF$ .
From $\angle SCF=\angle SBC_2=\angle SBC$ we have $\triangle SCF\sim\triangle SBC$ .
Therefore, $\frac{AD}{AC}=\frac{AD}{AB}=\frac{SF}{SB}=\frac{SF}{SC}\cdot\frac{SC}{SB}=\frac{CF^2}{CB^2}$ . $\square$

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#3 h Sep 15, 2024, 11:09 AM

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Apply Stewart.

$BC^2\cdot AD + AC^2\cdot BD = AB(CD^2+BD\cdot AD) = AC(FD^2+FC^2+BD\cdot AD)$
$BC^2\cdot AD + AC^2\cdot BD = AC\cdot FC^2 + AC(BD^2+BD\cdot AD)=AC\cdot FC^2 + AC\cdot BD (BD+AD)$

$BC^2\cdot AD =AC\cdot FC^2$

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#4 h Apr 19, 2025, 5:56 PM

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Why does S midpoint I am confused

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