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Question
If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see figure), prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle.
Solution
Given: In a circle AYDZBWCX, two chords AB and CD intersect at right angles.
To prove: arc CXA + arc DZB = arc AYD + arc BWC = Semi-circle.
Construction: Draw a diameter EF parallel to CD having centre M.
Proof: Since, CD || EF
arc EC = arc PD ...(i)
arc ECXA = arc EWB [Symmetrical about diameter of a circle]
arc AF = arc BF ...(ii)
We know that, ar ECXAYDF = Semi-circle
arc EA + arc AF = Semi-circle
⇒ arc EC + arc CXA = arc FB = Semi-circle ...[From equation (ii)]
⇒ arc DF + arc CXA + arc FB = Semi-circle ...[From equation (i)]
⇒ arc DF + arc FB + arc CXA = Semi-circle
⇒ arc DZB + arc C × A = Semi-circle
We know that, circle divides itself in two semi-circles, therefore the remaining portion of the circle is also equal to the semi-circle.
∴ arc AYD + arc BWC = Semi-circle
Hence proved.
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