Question

Question 4
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that $\angle ABC$ is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer 1

Given: Two equal chords AD and CE of a circle with centre O. They meet at B when produced.
To prove: $\angle ABC=\frac{1}{2}(\angle AOC-\angle DOE)$
Proof: let $\angle AOC=x,\angle DOE=y,\angle AOD=z$
$\angle EOC=z$ [Equal chords subtends equal angles at the centre]
$\therefore x+y+2z={360}^{\circ }$ [Angle at a point] ...(i)
$OA=OD\Rightarrow \angle OAD=\angle ODA$
$\therefore$ In $△$OAD, we have
$\angle OAD+\angle ODA+z={180}^{\circ }$ [by angle sum property of a triangle]
$\Rightarrow 2\angle OAD={180}^{\circ }-z$ $[\because \angle OAD=\angle OBA]$
$\Rightarrow \angle OAD={90}^{\circ }-\frac{z}{2}$ ...(ii)
Similarly, $\angle OCE={90}^{\circ }-\frac{z}{2}$ ...(iii)
$\Rightarrow \angle ODB=\angle OAD+\angle ODA$ [Exterior angle property]
$\Rightarrow \angle ODB={90}^{\circ }-\frac{z}{2}+z$ [From (ii)]
$\Rightarrow \angle ODB={90}^{\circ }+\frac{z}{2}$ ...(iv)
Also, $\angle OEB=\angle OCE+\angle COE$ [Exterior angle property]
$\Rightarrow \angle OEB={90}^{\circ }-\frac{z}{2}+z$ [From (iii)]
$\Rightarrow \angle OEB={90}^{\circ }+\frac{z}{2}$ ...(v)
Also, $\angle OED=\angle ODE={90}^{\circ }-\frac{y}{2}$ ...(vi)
From (iv), (v) and (vi), we have
$\angle BDE=\angle BED={90}^{\circ }+\frac{z}{2}-({90}^{\circ }-\frac{y}{2})$
$\Rightarrow \angle BDE=\angle BED=\frac{y+z}{2}$
$\Rightarrow \angle BDE+\angle BED=y+z$ ...(vii)
$\therefore \angle EBD={180}^{\circ }-(y+z)$ [by angle sum property of a triangle]
$\Rightarrow \angle ABC={180}^{\circ }-(y+z)$ ...(viii)
Now, $\frac{y-z}{2}=\frac{{360}^{\circ }-y-2z-y}{2}={180}^{\circ }-(y+z)$ ...(ix)
From (viii) and (ix), we have
$\angle ABC=\frac{x-y}{2}$. proved