Question

14. In the given figure, AD is the chord of the
larger of the two concentric circles and BC
is the chord of the smaller circle. Prove that
AB = CD.
​

Answer 1

Answer:

Drop OP ⊥ AD

∴ OP bisects AD

(Perpendicular drawn from the centre of a circle to a chord bisects it)

AP = PD ……………. (i)

Now, BC is a chord for the inner circle and OP ⊥ BC

∴OP bisects BC

(Perpendicular drawn from the centre of a circle to a chord bisects it)

BP=PC.........................(ii)

subtracting(ii) from (i)

AP-PB=PD-PC

=> AB=CD

Answer 2

Step-by-step explanation:

$\blue{ \bf{ \underline{QUESTION} \: : - }}$

In the given figure, AD is the chord of the

larger of the two concentric circles and BC

is the chord of the smaller circle. Prove that

AB = CD.

_________________________________

$\boxed{ \huge{ \bold{ Given}}}$

• AD is Chord of Large circles.

• BC is Chord of Small circles.

$\boxed{ \huge{ \bold{prove \: that}}}$

• AB = CD

_______

$\star{ \pink{ \underline{ \underline{Ｓｏｌｕｔｉｏｎ : - }}}}$

Drop

OE ⊥ AD

∴ OE bisects AD

( Perpandicular drawn from the centre of Circle to a Chord bisects it)

AD = PD ............. (1)

NOW,

BC is a chord for the inner circle and

OE ⊥ BC

∴ OE bisects BC

( Perpandicular drawn from the centre of a circle to a chord bisects it)

BP = PC ............. (2)

subtracting (2) from (1)

AP - BP = PD - PC

AB = CD

Proved *