Question

AB and CD are two chords of the circle with center O, which intersects at E.Prove that: $\angle AOC+\angle DOB=2\angle AEC$

Answer 1

AB and CD are two chords of the circle with center O, which intersects at E.
$TPT:\angle AEC=\frac{1}{2}(\angle COA+\angle DOB)$
Construction ∶ Join AC, AO, OC, BC and BD
Proof :
AC is a chord.
Therefore $\angle AOC=\mathrm{2......}\left(1\right)$ [angle subtended at center is double the angle subtended at circumference]
Similarly $\angle DOB=2\angle DCB$.....(2) [angles subtended by the chord BD]
Adding (1) and (2)
$\angle AOC+\angle DOB=2(\angle ABC+\angle DCB)......\left(3\right)$
In the triangle CEB,
$\angle AEC=\angle ECB+\angle CBE$ [Exterior angle is the sum of two opposite interior angles]
$\angle AEC=\angle DCB+\angle ABC....\left(4\right)$
From (3) and (4)
$\angle AOC+\angle DOB=2\angle AEC$