Question

can any one solve 8th question it is challenge for all. lets check who can solve this.

Answer 1

m (arc cxa)=mL COA...(1)central angle
m(arc dyb)=mL BOD..... (2)central angle
now
chords AB and CD intersecting internally
by property
mL AEC=1÷2[m(arc cxa) +m(arc dby)]
= 1÷2(L CAO+ L BOD).....from (1),(2)

Answer 2

CONSTRUCTION JOIN ,AO CO BC AND BD
AC IS THE CHORD
 SO ANGLE AOC=2 ANGLE ABC
ANGLE DOB =2 ANGLE DCB  ....1
ANDD AOB + DOC =2 [ABC +DCB].....2
IN TRIANGLE CEB
ANGLE AEC =NGLE ECB =NGLE CBE ...EXT ...3
ANGLE AEC=ANGLE DCB +ANGLE ABC...4
FROM 3AND 4
ANGLE AOC +ANGLE DOB =2]ANGLE AEC]
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