If XY and X'Y' are parallel tangent to a circle with Centre o and another tangent with point of contact c intersecting xy at a and X'Y' at B. Prove that angle son = 90°
Answers
In ΔOPA and ΔOCA,
OP = OC (Radii of the same circle)
AP = AC (Tangents from point A)
AO = AO (Common side)
∴ ΔOPA ≅ ΔOCA (SSS congruence criterion)
⇒ ∠POA = ∠COA … (i)
Similarly,
ΔOQB ≅ ΔOCB
∠QOB = ∠COB … (ii)
Since POQ is a diameter of the circle, it is a straight line.
∴ ∠POA + ∠COA + ∠COB + ∠QOB = 180 º
From equations (i) and (ii),
2∠COA + 2∠COB = 180º
⇒ ∠COA + ∠COB = 90º
⇒ ∠AOB = 90°
Given: XY is tangent to the circle at point P
X’Y’ is tangent to the circle at point Q
AB is tangent to the circle at point C
XY ∥ X’Y’
To prove: ∠AOB = 90°
Proof: First join OC such that OC is perpendicular to AB.
⇒ ∠ACO = 90° & ∠BCO = 90°
We know already that ∠OPA = ∠OQB = 90° (as XY and X’Y’ are tangents at P and Q respectively; and tangents are always perpendicular to any part of the circle)
Now taking ∆AOP and ∆AOC, observe that
OP = OC [∵ they are radius of the same circle; radius of a circle have equal lengths]
AP = AC [∵ length of tangents drawn from an external point to a circle are equal]
OA = OA [∵ they are same sides of the triangles]
Thus, by SSS-congruency ∆AOP ≅ ∆AOC.
⇒ ∠POA = ∠AOC [∵ corresponding parts of congruent triangles are equal] …(i)
Taking ∆COB and ∆BOQ, observe that
OQ = OC [∵ they are radius of the same circle and hence, are equal]
BQ = BC [∵ length of tangents drawn from an external point to a circle are equal]
OB = OB [∵ they are same sides of the triangle]
Thus, by SSS-congruency ∆COB ≅ ∆BOQ.
⇒ ∠COB = ∠BOQ [∵ corresponding parts of congruent triangles are equal] …(ii)
Breaking up line PQ into angles, we get
∠POA + ∠AOC + ∠COB + ∠BOQ = 180°
⇒ ∠AOC + ∠AOC + ∠COB + ∠COB = 180° [By equations (i) & (ii)]
⇒ 2 ∠AOC + 2 ∠COB = 180°
⇒ 2(∠AOC + ∠COB) = 180°
⇒ ∠AOC + ∠COB = 180°/2
⇒ ∠AOB = 90°
Hence, proved.
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