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Question

Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

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Solution

To prove :: AB||PT

Construction: join OA,OB,&OP

Proof:
OPPT [Radius is ⟂ to tangent through a point of contact]

OPT=90°

Since P is the midpoint of Arc APB

Arc(AAP)=arc(BP)

AOP=BOP

AOM=BOM

In ΔAOM&ΔBOM

OA=OB=r

OM=OM (Common)

AOM=BOM (proved above)

AOMBOM (by SAS congruence axiom)

AMO=BMO (C.P.C.T)

AMO+BMO=180°

AMO=BMO=90°

BMO=OPT=90°

But, they are corresponding angles. Hence, AB||PT

956769_427354_ans_ed65ee87a0304fbbbe073ee76ea9f580.png

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