Intersection of Three Chords

The butterfly theorem states that if $C$ is the center of a chord $AB$ in a circle, chords $DE$ and $FG$ pass through $C$ and $AB$ intersects $DG$ and $FE$ at $H$ and $I$ respectively, then $C$ is the midpoint of $IH$ . In this case we now know that $CI=CH=AC-AH=550-449=101$ , as desired.

By the Butterfly theorem, we see that C is the midpoint of HI.

Since C is the midpoint of HI, $CI=CH$ .

Since C is the midpoint of AB, $CA=CB=\frac{AB}{2}$

Thus, $CI=CH=CA-AH=\frac{AB}{2}-AH=\frac{1100}{2}-449=101$

Butterfly theorem states that $CI=CH$ , hence $CI= CH = AC-AH = 550 - 449 = 101$ .

To prove butterfly theorem, observe that $FI =CI \frac {\sin \angle FCA } {\sin \angle EFG }$ and $IE = CI \frac { \sin \angle ECA} {\sin \angle FED }$ . From power of a point, we have

$\begin{aligned} CI^2 \frac { \sin \angle FCA \cdot \sin \angle ECA} { \sin \angle EFG \sin \angle FED} &= FI \cdot IE \\ &= IA \cdot IB \\ &= (AC+CI)(BC-CI) \\ &= CA^2 - CI^2 \\ \end{aligned}$ Hence $CA^2 = CI^2 \left( 1 + \frac { \sin \angle FCA \cdot \sin \angle ECA} { \sin \angle EFG \sin \angle FED} \right)$ . Similarly for H, we have $CB^2 = CH^2\left( 1+ \frac { \sin \angle DCB \sin \angle BCG }{\sin \angle EDG \sin \angle FGD } \right)$ .

Since $\angle FCA = \angle BCG$ , $\angle ECA = \angle DCB$ , $\angle EFG = \angle EDG$ and $\angle FED = \angle FGD$ , the coefficients are the same, so $CH=CI$ .

we draw an imaginary circle.according to given condition we draw required segment.as given that c is mid point& we know that aline drawn from centre perpendicular to chord divides i.e. cis the mid point.