Question

E
Given : In a circle with centre B arc APC = arc DC
To Prove: Chord AC chord DE
Proof : (Fill in the blanks and complete t
proof.)
In A ABC and ADBE,
side AB side DB (........)
side .... side ........ (........)
ZABC = ZDBE measures of congrue
arcs
..A ABC = ADBE (.......)
:. chord AC chord DE (.....)​

Answer 1

Answer:

MATHS

In a circle with centre O, chords AB and CD intersect inside the circumference at E.Prove that <AOC+<BOD=2<AEC.

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ANSWER

In the given figure, the same arc AC subtends ∠AOC on the centre and ∠ABC on the other part.

⇒∠AOC=2∠ABC [Angle subtended on the centre of the circle is double the angle subtended on the other part of the circle by same arc]

Similarly, arc BD subtends ∠BOD and ∠DCB

⇒∠BOD=2∠DCB

On adding the above results, we get

∠AOC+∠BOD=2∠ABC+2∠DCB

=2(∠ABC+∠DCB)......(1)

In △ECB, by exterior angle property, we have

∠AEC=∠ECB+∠EBC

⇒∠AEC=∠DCB+∠ABC......(2)

On putting the value of (2) in (1), we get

∠AOC+∠BOD=2∠AEC

Hence proved.

solution