Question

Plz urgent fast reply plz plzFind the sum of the angles in the four segments exterior to a cyclic quadrilateral

Answer 1

Answer:

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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 ]÷••)—

Answer 2

hello dear...here is ur solution ❤️❤️


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




hope it helps u☺️☺️❤️❤️❤️