Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • ABCD is such a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that \angle CBD + \angle CDB =\frac{1}{2}\angle BAD.

ABCD is such a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that  \angle CBD + \angle CDB =\frac{1}{2}\angle BAD.

Answers (1)

Given: A circle passing through points B, C, D and ABCD is a quadrilateral having centre at A.

To prove: \angle CBD +\angle CDB =\frac{1}{2}\angle BAD

Construction: Join AC & BD

Proof:

We know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle

Arc DC subtends \angleDAC at the centre and \angleDBC at a point B in the remaining part of the circle.

\therefore \angleDAC = 2\angleCBD … (i)

Similarly, arc BC subtends ÐBAC at the centre and ÐBDC at a point D in the remaining part of the circle.

\therefore \angleCAB = 2 \angleCDB … (ii)

On adding equations (i) and (ii)

We get

\angleDAC + \angleCAB = 2 \angleCBD + 2 \angleCDB

\angleDAB = 2 (\angleCBD + \angleCDB)

\angle CBD +\angle CDB =\frac{1}{2}\angle BAD

Hence proved

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question