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Chord of the Bigger Circle Is Bisected at the Point of Contact with the Smaller Circle

Question 6 If...

Question

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.

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Solution

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
$\Rightarrow$ arc EC + arc CXA + arc FB = semi-circle [from Eq. (ii)]
$\Rightarrow$ arc DF + arc CXA + arc FB = Semi-circle [from Eq. (i)]
$\Rightarrow$ arc DF + arc FB + arc CXA = Semi-circle
$\Rightarrow$ 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

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Q. Question 8
In the figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that

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In the given figure, $A B$ and $C D$ are two chords of a circle, intersecting each other at a point $E .$ Prove that $∠ A E C = \frac{1}{2}$ (angle subtended by arc $C X A$ at the centre $+$ angle subtended by arc $D Y B$ at the centre).

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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.

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.