Two circles are externally tangent to each other at a point C. A common external tangent (AB) is drawn to them, where A and B are points of tangency. Prove that $∠ A C B = π / 2$

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Solution

In the triangle $△ A M C$ , let $∠ A M C = x ⟹ ∠ A C M = 90 - \frac{x}{2}$

In the triangle $△ B N C$ , let $∠ B N C = x ⟹ ∠ B C N = \frac{x}{2}$

Thus, $∠ A C B = 180 - (∠ A C M + ∠ B C N) = 180 - 90 = 90^{0} = \frac{π}{2}$

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Two circles are externally tangent to each other at a point C. A common external tangent (AB) is drawn to them, where A and B are points of tangency. Prove that ∠ACB=π/2

In the triangle △AMC, let ∠AMC=x⟹∠ACM=90−x2In the triangle △BNC, let ∠BNC=x⟹∠BCN=x2Thus,∠ACB=180−(∠ACM+∠BCN)=180−90=900=π2

Two circles are externally tangent to each other at a point C. A common external tangent (AB) is drawn to them, where A and B are points of tangency. Prove that $∠ A C B = π / 2$

In the triangle $△ A M C$ , let $∠ A M C = x ⟹ ∠ A C M = 90 - \frac{x}{2}$

In the triangle $△ B N C$ , let $∠ B N C = x ⟹ ∠ B C N = \frac{x}{2}$

Thus, $∠ A C B = 180 - (∠ A C M + ∠ B C N) = 180 - 90 = 90^{0} = \frac{π}{2}$