The line segment joining the midpoints of two parallel chords of a circle passes through the centre. AB and CD are two equal chords of a circle with centre O. If AB and CD, on being produced meet at a point P outside the circle, Prove that:- (i) PA=PC (ii) PB=PD
Answers
Answer:
Join points B and C as shown in the figure.
∠M=∠N=90
o
(Segment joining the midpoint of the chord and center of the circle is perpendicular to the chord.)
Now, in △OMB and △ONC,
∠BOM=∠CON ....Vertically opposite angles
∠OMB=∠ONC ...Each 90
o
OB=OC ...radii of the same circle
∴△OMB≅△ONC ...SAA test
∴∠OBM=∠OCN ....C.A.C.T
∴∠ABC=∠BCD ...Since A−M−B and C−N−D
These are nothing but alternate angles for the chords AB and CD respectively
Answer:
Given: M and N are midpoints of chord AB and chord CD respectively.
MN passes through the center of the circle O.
To prove: chord AB∥ chord CD
Solution:
Join points B and C as shown in the figure.
∠M=∠N=90
o
(Segment joining the midpoint of the chord and center of the circle is perpendicular to the chord.)
Now, in △OMB and △ONC,
∠BOM=∠CON ....Vertically opposite angles
∠OMB=∠ONC ...Each 90
o
OB=OC ...radii of the same circle
∴△OMB≅△ONC ...SAA test
∴∠OBM=∠OCN ....C.A.C.T
∴∠ABC=∠BCD ...Since A−M−B and C−N−D
These are nothing but alternate angles for the chords AB and CD respectively.
Hence, chord AB∥ chord BC.