In an isosceles triangle the angles at the base are equal. |
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Let ABC be an isosceles triangle in which side AB is equal to side AC; |
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then angle ABC is equal to angle ACB. |
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Let us conceive of this triangle as two triangles -- the triangles ABC |
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and ACB -- and let us argue as follows. |
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Since AB is equal to AC, and AC to AB, |
(Hypothesis) |
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then the two sides BA, AC are equal to the two sides CA, AB respectively; |
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and angle BAC is equal to angle CAB, because it is the same. |
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Therefore those angles are equal that are opposite the equal sides: (S.A.S.) |
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angle ABC, opposite side AC, is equal to angle ACB, opposite the equal side AB. |
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Therefore, in an isosceles triangle the angles at the base are equal. |
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Which is what we wanted to show. |