Question

$\overline{AB}$ and $\overline{CD}$ are chords of a circle with radius r. AB=2 CD and the perpendicular distance of $\overline{CD}$ from the centre is twice perpendicular distance of $\overline{CD}$ from the centre is twice perpendicular distance of $\overline{AB}$ from the centre. Prove that $r=\frac{\sqrt{5}}{2}CD$.

Answer 1

According to the problem $AB=2CD$.......(1) and $OF=2OE$......(2).

As $\Delta$ OBE $\cong \Delta$AOE, we have $AE=EB=\frac{1}{2}AB$......(3).

Also $\Delta$ ODF $\cong \Delta$OCF, we have $DF=CF=\frac{1}{2}CD$......(4).

From (2) we have

$OF=2OE$

or, $\sqrt{r^2-D{F}^2}=2\sqrt{r^2-E{B}^2}$

or, $r^2-D{F}^2=4(r^2-E{B}^2)$

or, $r^2-\frac{C{D}^2}{4}=4(r^2-C{D}^2)$ [Since $EB=\frac{AB}{2}=CD$]

or, $3r^2=\frac{15C{D}^2}{4}$

or, $r=\frac{\sqrt{5}}{2}CD$.