Q27 of 63 Page 156

In figure, AB and CD are two chords of a circle which intersect at E. Prove that (angle subtended by arc CXA at centre + angle subtended by arc DYB at centre).


Given AB and CD are two chords of the circle with centre O which intersects at E.


We have to prove that ∠AEC = 1/2 (angle subtended by arc CXA at centre + angle subtended by arc DYB at centre).


Construction:


Join AC, BC and BD.


Proof:


AC is a chord.


We know that angle subtended at center is double the angle subtended at circumference.


⇒∠AOC = 2∠ABC … (1)


⇒∠DOB = 2∠DCB … (2)


Adding (1) and (2),


⇒∠AOC + ∠DOB = 2(∠ABC + ∠DCB) … (3)


Consider ΔCEB,


We know that the sum of two opposite interior angles is the exterior angle.


⇒∠AEC = ∠ECB + ∠CBE


⇒∠AEC = ∠DCB + ∠ABC … (4)


From (3) and (4),


∴ ∠AOC + ∠DOB = 2 (∠AEC)


Hence proved.


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