Circles Γ1 and Γ2 intersect at 2 distinct points A and B. A line l through A intersects Γ1 and Γ2 at points C and D, respectively. Point E is the intersection of the tangent to Γ1 at C and the tangent to Γ2 at D. If ∠CBD=30∘, what is the measure (in degrees) of ∠CED?
To solve this problem, we will use some properties of circles and angles.
Given that Γ1 and Γ2 intersect at points A and B, we can see that triangles ABC and ABD are similar. This is because they share a common angle, ∠CBA = ∠DBA, and have an identical angles, ∠ACB = ∠ADB, due to tangents being perpendicular to radii at the point of contact.
We also know that ∠CBD = 30∘. Since triangles ABC and ABD are similar, we have ∠CAB = ∠DAB = ∠CBD = 30∘.
Now, let's consider the triangle CED formed by the tangents to Γ1 and Γ2 at points C and D, respectively. We are asked to find the measure of ∠CED.
Since line l is a tangent to Γ1 at C, by the properties of tangents, ∠CDE = 90∘. This is because the angle formed by the tangent and the radius at the point of contact is always 90 degrees.
Now consider ∠BAC in triangle ABC. Since Γ1 and Γ2 intersect at points A and B, we know that ∠BAC = ∠CAB = 30∘.
Using the fact that the sum of the angles in a triangle is 180∘, we can find ∠CED.
Since ∠CDE = 90∘, and ∠CED + ∠CDE + ∠C = 180∘, we have:
∠CED + 90∘ + 30∘ = 180∘
Simplifying the equation:
∠CED = 180∘ - 90∘ - 30∘
∠CED = 60∘
Therefore, the measure of ∠CED is 60 degrees.