**Q) In the given figure, AB, BC, CD and DA are tangents to the circle with centre O forming a quadrilateral ABCD. Show that angle AOB+ angle COD = 180 ^{0} **

**Ans: **

Let’s draw a diagram and connect O with all vertices of Quadrilateral ABCD and al touch points on its circumference:

Let’s start with Δ AOP and Δ AOS:

OA = OA (common arm)

OP = OS (radii of circle)

AP = AS (tangents to a circle)

∴ Δ OAP Δ OAS

We know that in two congruent triangle, two corresponding parts are always equal (CPCT)

∴ ∠ AOS = ∠ AOP = w

Similarly, Δ POB Δ QOB and ∠ POB = ∠ QOB = x

Similarly, Δ QOC Δ ROC and ∠ QOC = ∠ ROC = y

Similarly, Δ ROD Δ SOD and ∠ ROD = ∠ SOD = z

Next, we know that sum of all angles at a point is always 360

∴ ∠ AOS + ∠ AOP + ∠ POB + ∠ QOB + ∠ QOC + ∠ ROC + ∠ ROD + ∠ SOD = 360^{0}

∴ w + w + x + x + y + y + z + z = 360^{0}

∴ 2 (w + x + y + z) = 360^{0}

∴ (w + x + y + z) = 180^{0}

∴ (w + x) + (y + z) = 180^{0}

**∴ ∠ AOB + ∠ COD = 180 ^{0}**

**Hence Proved !**

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