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P-2 Circle — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsP-2 Circle4 Marks
Question
In figure 3.102, two circles intersect each other at points A and E. Their common secant through E intersects the circles at points B and D. The tangents of the circles at points Band D intersect each other at point C.
Prove that $\square$ABCD is cyclic.
✓

Answer

We join A to B and A to D and A to E

As BC is a tangent at B, we have
∠ CBD = ∠BAE ………….(1)
As CD is a tangent at D, we have
∠ CDB = ∠DAE ………….(2)
In ΔBCD, we have
⇒ ∠CBD + ∠CDB + ∠ BCD = 180° (Sum of all angles of a triangle)
⇒ ∠BAE + ∠DAE + ∠BCD = 180° (From (1) and (2))
⇒ ∠BAD + ∠BCD = 180° (∠BAE + ∠DAE = ∠BAD)
In quadrilateral ABCD,
We have ∠A + ∠C = 180° (Proved above)
⇒ ∠A + ∠B + ∠C + ∠D = 360°
⇒ ∠B + ∠D + 180 = 360
⇒ ∠B + ∠D = 180
Therefore, opposite angles of the quadrilateral sum to 180. Hence ABCD is a cyclic quadrilateral.

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