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We are given isosceles triangle ABC, with vertex angle A.(If A is not the vertex angle, then AB=BC and the proposition is false.) Angle A is 20 degrees. Prove 2BC<AB<3BC.
Drop an altitude from A. In an isosceles triangle, the altitude is also the perpendicular bisector of the base as well as the angle bisector of the vertex angle. Label the point of intersection of the altitude with the base as M. Then we have created right triangle AMB.
In triangle AMB, angle BAM measures 10 degrees and angle B measures 80 degrees. Then using trigonometric ratios we have:
`cos80=(BM)/(AB)` or `BM=AB cos80=>BC=2ABcos80~~.347AB`
Then `.695AB=2BC<AB<3BC=1.04AB` as required.
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