Prove that the sum of the angles formed in the four segments exterior to a cyclic quadrilateral by the sides is equal to six right angles
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Step-by-step explanation:
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Step-by-step explanation:
Let ABCD is a Cyclic quadrilateral
∠P,∠Q,∠R and ∠S are angles in the four external segments
To Prove : ∠P,∠Q,∠R,∠S =6 Right Triangles
Contruction : Join SB and SC
From the Figure APBS is Cyclic quadrilateral
∠ASB + ∠P = 180°...................... (1)
Similarly BQCS is Cyclic quadrilateral
∠CSB + ∠Q = 180°...................... (2)
CRDS is Cyclic quadrilateral
∠CSD + ∠R = 180°...................... (3)
Now, Adding 1, 2 and 3,we get
∠ASB + ∠P + ∠CSB + ∠Q + ∠CSD + ∠R = 180°+ 180° + 180°
∠P + ∠Q + ∠R +∠ASB + ∠CSB +∠CSB = ∠S
⇒∠P + ∠Q + ∠R + ∠S = 3 x 180°
⇒∠P + ∠Q + ∠R + ∠S = 3 x 2 x 90°
⇒∠P + ∠Q + ∠R + ∠S = 6 x 90°
⇒∠P + ∠Q + ∠R + ∠S = 6 x Right angles
—(••÷[ тнχ ]÷••)—
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