Question 6
If two chord AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle.
If two chord AB and CD of a circle AYDZBWCX intersect at right angles, then prove that arc CXA + arc DZB = arc AYD + arc BWC = semi-circle.
Given in a circle AYDZBWCX, two chords AB and CD intersect at right angles
To prove that:
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 FD............(i)
Also, arc ECXA = arc EWB [symmetrical about diameter of a circle]
and arc AF = arc BF............(ii)
We know that, arc ECXAYDF = Semi-circle
Arc EA + arc AF = Semi-circle
⇒ arc EC + arc CXA + arc FB = semi-circle [from Eq. (ii)]
⇒ arc DF + arc CXA + arc FB = Semi-circle [from Eq. (i)]
⇒ arc DF + arc FB + arc CXA = Semi-circle
⇒ arc DZB + arc CA = 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
Given in a circle AYDZBWCX, two chords AB and CD intersect at right angles
To prove that:
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 FD............(i)
Also, arc ECXA = arc EWB [symmetrical about diameter of a circle]
and arc AF = arc BF............(ii)
We know that, arc ECXAYDF = Semi-circle
Arc EA + arc AF = Semi-circle
⇒ arc EC + arc CXA + arc FB = semi-circle [from Eq. (ii)]
⇒ arc DF + arc CXA + arc FB = Semi-circle [from Eq. (i)]
⇒ arc DF + arc FB + arc CXA = Semi-circle
⇒ arc DZB + arc CA = 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
