Question

plz slove this 9th class ☺️​

Answer 1

Answer:

plz mark it as brainliest

Explanation:

8th

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

We know that OB⊥AB

From the figure we know that D is the midpoint of AB

We get

AD=BD

We also know that O is the midpoint of BC

We get

OC=OB

Consider △ABC

Using the midpoint theorem

We get OB∥AC and

OD= 1/2 ×AC

By cross multiplication

AC=2×OD

Therefore, it is proved that AC∥DO and AC=2×OD

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9th

Consider △OEP and △OFP

We know that

∠OEP=∠OFP=90

o

OP is common i.e. OP=OP

From the figure we know that OP bisects ∠BPD

It can be written as

∠OPE=∠OPF

By ASA congruence criterion

△OEP≅△OFP

OE=OF (c.p.c.t)

We know that AB and CD are equidistant from the centre

So we get

AB=CD

Therefore, it is proved that AB=CD....

hope it helps you