Question

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Answer 1

Question

In the figure XY and X'Y' are two parallel tangents to a circle with centre O and and another tangent AB with point of contact C interesting XY at A and X'Y' at B prove that ∠AOB=90  °

Answer:

⇒OP⊥ xy (tangent  radius)

⇒ OC ⊥ AB

In △OPA and △ OCA

⇒ ∠OPA =  ∠OCA = 90 °

⇒ OP = OC (radii)

⇒ AP = AC (tangents from an external point)

△ OPA  ≅△OCA (SAS)

∴ ∠ 1 =∠ 2 (CPCT)

∴ ∠2 =1/2 ∠PAC

Similarly,∠ 3 = ∠4

⇒ ∠ 3 = 1/2∠ QBC

xy ||x'y' and AB is transversal

⇒ ∠ PAB +∠ QBA = 180o (interior angles on same side of transversal)

⇒ Or ∠PAC = ∠QBC = 180 °

⇒  ∠ 2 +∠ 3 = 90 °

In △ OAB,

⇒ ∠AOB + ∠2 +∠ 3 = 180 °

⇒ ∠ AOB = 90 °

Hence proved.