Question

given figure O is the centre of a circle if a b and CD are two equal chords of the circle op perpendicular to a b and 4q perpendicular to CD prove that P B equals to QC ​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

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It is given that

AB = AC

Dividing the equation by 2

We get

½ AB = ½ AC

Perpendicular from the centre of a circle to a chord bisects the chord

MB = NC ……. (1)

We know that the equal chords are equidistant from the centre of the circle

OM = ON and OP = OQ

Subtracting both the equations

OP – OM = OQ – ON

So we get

PM = QN ……… (2)

Consider △ MPB and △ NQC

We know that

∠PMB = ∠QNC = 90°

By SAS congruence criterion

△ MPB ≅ △ NQC

PB = QC (c. p. c. t)

Therefore, it is proved that PB = QC