Question

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Answer 1

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AB and DC are two chords of a circle, centre O. If AB and DC meet at P and OP bisects <APC internally, how do you prove that AB = DC?

In triangles OBP and OCP,

☆☞OB=OC [being radii of the same circle]

☆☞<BPO = <CPO [as OP bisects <APC]

☆☞OP is common.

Therefore triangles OBP and OCP are congruent. So BP = CP.

In triangles AOP and DOP, OA=OD [being radii of the same circle]

<APO = <DPO [as OP bisects <APC]

OP is common.

Therefore triangles AOP and DOP are congruent. So AP = DP.

But BP = CP, so AP-BP (which is AB) =DP-CP (which is DC).

$\boxed{\boxed{Thus AB = DC. Proved.}}$

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Answer 2

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step-by-step explanation:

Given,

AB and DC are two chords of a circle having centre at O.

Now,

In triangles OBP and OCP, 

OB=OC ( radii of the same circle )

angle BPO = angle CPO ( °.° OP bisects <APC )

OP is common.

Therefore,

∆ OBP and ∆OCP are congruent.

So,

BP = CP ( by C.P.C.T)

NOw,

In ∆ AOP and ∆DOP,

OA=OD ( radii of the same circle)

angle APO = angle DPO

( °.° OP bisects angle APC )

also,

OP is common.

Therefore,

∆ AOP and ∆DOP are congruent.

So,

AP = DP.

But,

BP = CP,

so,

(AP-BP) = ( DP-CP)

but,

AP -BP = AB

DP-CP = DC

therefore,

AB = CD

hence,

proved✍️✍️