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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Answer:
Given:
A circle with Centre O, P is the midpoint of Arc APB. PT is a tangent to the circle at P.
To Prove:
AB || PT
Construction: join OA ,OB, & OP
Proof: OP ⟂PT
[Radius is ⟂ to tangent through point of contact]
∠OPT= 90°
Since P is the midpoint of Arc APB
Arc AAP =arc BP
∠AOP = ∠BOP
∠AOM= ∠BOM
In ∆ AOM & ∆BOM
OA= OB= r
OM = OM (Common)
∠AOM= ∠BOM (proved above)
∠AOM≅∠BOM (by SAS congruency axiom)
∠AMO = ∠BMO (c.p.c.t)
∠AMO + ∠BMO= 180°
∠AMO = ∠BMO= 90°
∠BMO = ∠OPT= 90°
But, they are corresponding angles. Hence, AD||PT
HOPE THIS WILL HELP YOU... .
fanoffootball7:
where is M in the cirle
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