proving relative lengths on a secant

This image matches the description in the question (note that, in violation of what I consider to be conventional, $O$ is not the centre of the circle but the midpoint of $AD$). I add a perpendicular to $AD$ from $O$ which intersects $AC$ at $G$, and lines $OF$ and $DG$ which intersect at $H$.

The key idea is to show that $AOG$, $DOG$, $DFG$ and $CDF$ are all congruent. Then since the area of a triangle is a half of the base times the height, $$\tfrac{1}{2} \overline{AF} \cdot \overline{DF} = 3 \cdot \tfrac{1}{2} \overline{CF} \cdot \overline{DF}$$

One argument which uses fairly basic lemmata and certainly makes no mention of similar triangles is as follows:

1. Angle $ACD = 60^\circ$ because it's an angle of an equilateral triangle.
2. Angle $CAD = 30^\circ$ because AD is a bisector.
3. $\overline{AO} = \overline{OD}$ by definition of $O$.
4. Angles $AOG = DOG = 90^\circ$ by definition of $G$.
5. Angle $AGO = 60^\circ$ by sum of angles in a triangle.
6. $AOG$ is congruent to $DOG$ by side-angle-side.
7. Angle $ODG = 30^\circ$ and $OGD = 60^\circ$ in consequence.
8. Angle $DGF = 60^\circ$ by sum of angles on a line.
9. $AO$ and $OF$ are radii of the circle, so $AOF$ is isosceles and angle $OFA = OAF = 30^\circ$.
10. Angle $FOG = 30^\circ$ by sum of angles in a triangle.
11. Angle $DOF = 60^\circ$ by e.g. sum of angles on a line.
12. $OD$ and $OF$ are radii of the circle, so $DOF$ is isosceles and angle $ODF = OFD$. But since angles in a triangle sum to $180^\circ$ and $DOF = 60^\circ$ that means that $ODF = OFD = 60^\circ$.
13. The angles of $OGD$ and $DFG$ are all known and they share edge $DG$, so we have enough to show that they're congruent.
14. Angle $DFC = 90^\circ$ by sum of angles on a line.
15. Angle $FDC = 30^\circ$ by e.g. sum of angles in a triangle.
16. The angles of $DFG$ and $DFC$ are known and they share edge $DF$, so we have enough to show that they're congruent.

posted almost 3 years ago

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3y ago

Peter Taylor‭

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