Question

in the figure O is the centre of a circle. prove angleAOC=angleAFC+angleAEC

Answer 1

choose the attachment

Answer 2

$\huge \bf{ \purple{ \underline{ \underline{Solution}}}}$

In this figure

$\bf \angle APC= \angle \: AXC$

(Angles on same arc)

$\bf \implies \: \angle \: EPC= \angle \: EXA---(1)$

(AE and EC are line)

Now ,in quadrilateral EPEX,

$\bf \angle E + \angle P+ \angle \: X+ \angle \: F=360 \degree$

$\bf \implies \: 2 \angle \: P+ \angle \: E+ \angle \: P=360 \degree$

$\bf \angle \: AOC+ \angle \: E+ \angle \: F=360 \degree$

$\: \: \: \: \: \: \bf \: \: \: \: \: \: \: ( \angle \: \:APC= \angle \: AOC)$

$\bf \implies \angle \: E+ \angle \: F=360 \degree- \angle \: AOC$

$\bf \implies \angle \: E+ \angle \: F= \angle \: AOC(reflex)$

$\bf \implies \: \angle \: AOC = \angle \: AFC+ \angle \:AEC$

Step-by-step explanation:

The figure is in the attachment.