Q25 of 63 Page 156

If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see figure), then prove that arc CXA + arc DBZ = arc AYD + arc BWC = a semicircle.


Given AB and CD are two chords of a circle AYDZBWCX such that AB ⊥ CD.


Let the chords intersect at O.


We have to prove that arc CXA + arc DBZ = arc AYD + arc CWB = semicircle.


Construction:


Join AD, DB, AC and BC.


Proof:


Angle subtended by chord AC is ∠CBA and angle subtended by chord BD is ∠BCD.


⇒∠CBA + ∠BCD = 180° - ∠COB


= 180° - 90°


= 90°


Since angle subtended in a semicircle is 90°,


⇒ arc CXA + arc DBZ = semicircle


Similarly arc AYD + arc CWB = semicircle


∴ arc CXA + arc DBZ = arc AYD + arc CWB = semicircle


Hence proved.


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