Answer :

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.

Rate this question :

In figure, if <spRajasthan Board Mathematics

If in any circle Rajasthan Board Mathematics

In figure, if AOBRajasthan Board Mathematics

In figure, if <spRajasthan Board Mathematics

In figure, if <spRajasthan Board Mathematics

If in congruent cRajasthan Board Mathematics

The degree measurRajasthan Board Mathematics

The circumcentre Rajasthan Board Mathematics

If an equilateralRajasthan Board Mathematics

The degree measurRajasthan Board Mathematics