Question
Asked Sep 24, 2019
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|Let ABC be an isosceles triangle, where AB= AC. Suppose that M is a point on the side
AB such that AB MB = AM2 . Suppose also that AM BC. Prove that the angle ZABC is
twice ZBAC. [Hint: Consider the circumcircle O of AMC. Show that BC is tangent to O
Use this to show that ZCMB= ZCBM . You are encouraged to use propositions from Books 1
| 2, & 3 from Euclid's Elements.] Draw a picture to help support your proof.
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|Let ABC be an isosceles triangle, where AB= AC. Suppose that M is a point on the side AB such that AB MB = AM2 . Suppose also that AM BC. Prove that the angle ZABC is twice ZBAC. [Hint: Consider the circumcircle O of AMC. Show that BC is tangent to O Use this to show that ZCMB= ZCBM . You are encouraged to use propositions from Books 1 | 2, & 3 from Euclid's Elements.] Draw a picture to help support your proof.

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Expert Answer

Step 1

To show that (under the given conditions) that angle A = 36 and (hence B=72=2A)

Step 2

We may take AB = AC =1 and AM = BC=x. The given condition yields a quadratic equation for x.

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АМ 2-АВ.ВM implies, х^2 3D1.(1-х) х х^2+x-1-0. ме 1-х C х

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Step 3

Solve for x.&...

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x2 x-10 -15 so, x 2 5-1 2

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