Our Geometry Project Exterior Angle Theorem Part 1 Converse of The Isosceles Triangle Theorem What we need : We know that if the sides of the triangles are the same, than the base angles are congruent, but do we know that if the base angles are equal than are the sides are equal?.... Construction Given b, an Angle, and a+c Proof : We are given a segment with length b, an

angle a,whose measure is α (alpha), and the sum of the other 2 sides, a+c. Copy alpha at A, by construction. Construct AD so that it equals a+c. Connect C to D by postulate 3&4. Construct <D at point C so that its side is CD. Call the intersection point of the new side of the angle and ∆ADE.

We claim that ∆AEC is desired a triangle. Period: 4 / Mr. Patrick

Assume BC>AB. Hence, we can extend AB by postulate 5

such that BD=BC. Connect D to C by postulate 3&4. Take ∆BDC, which is isosceles by construction and definition. <BDC=<BCD by the isosceles triangle theorem.b =a , because a is apart of b and apart is smaller than a whole. a >b by the ______________. This is impossible because if b >a and a >b , we get a contradiction (aka: DISASTER) since b =b and a =a . Hence, AB=CB is impossible, and as a result, AB=AC. Thus, if base angles of and triangle are = than the sides opposite the base angles are congruent. Proof: We are given ∆ADC with extended side at DA, forming an exterior angle, a. Connect C to the midpoint of AD, by construction. AE=DE by definition of bisector or midpoint. Extend CE so that CE=FE by postulate 5 and construction. Connect F to A by postulate 3&4. <FEA=<DFC by vertical angles theorem. By SAS, ∆FAE=∆CDE. By CPCTC, <CDE=<FAE. Extend FA by postulate 5 & construction <FAE=<GAH by V.A. theorem Since <EDC=<FAE and <FAE=<GAH, <EDC=<GAH because 2 things equal to a third are equal to each other. <GAC = <GAH by part/whole. ~ ~ ~ ~ ~ ~ ~ ~ ~ ≅ Our Geometry Project By: Elaina Cavada, Arriane Gatlabayan, & Robylene Seapno Exterior Angles Theorem Part : 2 Proof: We are given ∆ABC with extended side AD that forms exterior <a. Connect A to midpoint of BC by definition of construction. BE=CE by definition of midpoint. Extend AE so that AE=FE by postulate 5. Connect B to F by postulate 3&4. <AEC=<FEB by vertical angles theorem. By SAS ∆BEF=∆CEA. By CPCTC <ECA=<EBF. Since <FBE is a part of <a, and is congruent to <c, <a has to be larger than <c. That's our project How Everything Ties Together Proof: Take ∆ABC. We're given that the base angles are congruent (<A,<C). Suppose that AB=CB. 1 1 1 2 2 1 1 1 1 2 2 2 2 We get that in any triangle, its exterior angles is larger than the interior angle next to it. For the construction (given side b and a+c) problem, we needed to know that if a triangles base angles are equal than its sides opposite from them are equal. For the converse of the isosceles triangle theorem problem, we needed to know that a triangles interior angles are smaller than its exterior angles. For E.A.T. (exterior angle theorem), we solved the converse question which lead on to solve the construction(given side b and a+c) problem. / What we need: We need to know why b =a . we have an idea, but we need a reason. WTS: construct triangles with givens WTS: if base angles are =then the sides opposite the angles are =. WTS: its exterior angle(a) is larger than its interior angle(b) WTS: its exterior angle(a) is larger than its interior angle(c) That's our project part 1 part 2

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# Geometry project

Three proofs

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