Problem 14
Let ABC be a right angled triangle. A circle Г have side AC as its diameter meets hypotenuse AB at point E. A tangent line to Г at point E meets side BC at point D. Prove that a triangle BDE is isosceles.
Problem 14
Let ABC be a right angled triangle. A circle Г have side AC as its diameter meets hypotenuse AB at point E. A tangent line to Г at point E meets side BC at point D. Prove that a triangle BDE is isosceles.
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14. Let be a right triangle: .
A circle has side as its diameter, meets hypotenuse at
A tangent line to at meets side at point
Prove that is isosceles.Code:F o \ θ'\ A o * \ | θ * *\ E | θ o | * * * | * \ θ ' * | * *\ * O * * \ * | * \ * | \ * | * \ * | * \ * | * \ θ' * C o * - - - - - - - o - - - - - - - - - - - - - - o B D
Draw radius
Since is isosceles.
. .
Since is tangent at
. . Hence, and are complementary.
. . Let
and are vertical angles: .
In and are complementary.
. . Hence: .
In
Therefore, is isosceles.