If two equal chord of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segment of the outer chord.
Answers
Answered by
2
Answer:
the segments of one chord are equal to corresponding segment of the outer chord.
Attachments:
Answered by
2
Answer:
Step-by-step explanation:
Given AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.
To prove : AP = PD and PB = CP.
Construction: Draw OM perpendicular to AB and ON perpendicular CD. Join OP.
AM = MB = 1/2AB (Perpendicular bisecting the chord)
CN = ND = 1/2CD (Perpendicular bisecting the chord)
AM = ND and MB = CN (As AB = CD)
In triangle OMP and ONP, we have,
OM = MN (Equal chords are equidistant from the centre)
<OMP = <ONP (90⁰)
OP is common. Thus triangle OMP and ONP are congruent (RHS).
MP = PN (cpct)
So, AM + MP = ND + PN
or, AP = PD (i)
As MB = CN and MP = PN,
MB - MP = CN - PN
= PB = CP (ii)
Hope that helps !!
Similar questions