This is the Euclid's Proposition 5 from Book 1
"Isosceles" essentially means "equally-sided" which is from latin which in turn is from greek is-o-skelos ( read like spanish ). It means "equally - legged". What do you think the first "isoskelos" figures would look like? Like a pyramid? The third side is called "a base".
1.
A usual depiction of an isosceles triangle like a pyramid or a letter A is a subjective choice. Any triangle build upon two radiuses of a circle or a sphere is an isosceles one and it can take any orientation in space. |
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2. But we will use a usual drawing. Let ABC be an isosceles triangle where sides AB=AC. Need to proof that β equals γ. Plan of the proof. So far we have one theorem on equality of triangles. It is Proposition 4 which says that two triangles are equal if two corresponding sides and an angle between them are equal. Therefore if we construct two triangles which are equal and ABC is a part of both of them and if β and γ are parts of them and have a certain symmetry we may be able to conclude that β equals γ. |
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3.
In his proof Euclid introduces an algebraic operation of difference on lines and angles. A "whole" may contain two parts and if there are two "wholes" and they are equal, and one part in each "whole" is equal", the reminding parts are equal. In other words:
It works for line sections as well as for angles. And in the proof it is applied once to each. |
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4. Let's prolong the line AB (Euclid Postulate 2) and choose an arbitrary point F on it as it shown. A choice is made here, there could be another choice. Let prolong line AC and construct a point G such that AG=AF (Euclid Proposition 3). From 3 it follows that BF=CG. So, we did it for line sections. |
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5. Let's construct lines BG and CF (Euclid Postulate 1). We have constructed triangles ABG and ACF. From here it follows that CF=BG and angles AFC=AGB. From 4, we know that BF=CG. Therefore triangles BCG=CBF by (Euclid Proposition 4 ). |
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6. A side note. ABC is a part of ABG. It is how to construct a triangle which includes an initial triangle as a part with prolongation of the side and taking a point on it outside of AC. Another case would be if a point G is taken inside AC. The drawing illustrates that a line a triangle side coinsides with is a place of points forming a family of triangles where each one is a part of another. According with the plan two triangles ACF and ABG are made and ABC is a part of each, it belongs to each. |
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7. Considering 3, from the construction, for angles: ABG = CBG + ABC and from 5, ABG=ACF, and CBG=BCG. There fore ABC = ACB, or β = γ. So we did 3. for angles |
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8. From 5 follows that CBF = BCG, or "the angles under the base in isosceles triangles are equal". |
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9. Besides being formed by radiuses of a circle or a sphere ( see 1. ), isosceles triangle can be encountered, when two circles with the same radius intersect. A family of such circles drawn in the same plane around a pair of points and corresponding triangles is shown on the picture. |
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Problems |
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Problem 1 Prove that angles AFG and AGF equal. Solution: Triangle AFG is isosceles, there fore .. Q.E.D. |
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Problem 2 Construct an isosceles triangle ABC so that the base BC is parallel to the edge of the sheet of paper. You can use other theorems. |
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Problem 3* Show (or prove) that it is impossible to prove that angles ABC=AFG based on first 5 theorems of Euclid. What is missed? |