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Proving Triangles Similar Blind Sequencing Activity ̅̅̅̅ ≅ 𝐺𝐻 ̅̅̅̅, 𝐽𝐾 ̅̅̅ ≅ 𝐾𝐿 ̅̅̅̅, ∠𝐹 ≅ ∠𝐽 Given: 𝐹𝐺 K G Prove: ∆𝐹𝐺𝐻 ~ ∆𝐽𝐾𝐿 J F L H ̅̅̅̅ ∥ 𝐾𝐽 ̅̅̅ Given: 𝐹𝐺 K G Prove: ∆𝐹𝐺𝐻 ~ ∆𝐾𝐽𝐻 H F J 𝐹𝐺 Given: 𝐽𝐾 = 𝐺𝐻 𝐾𝐿 , ∠𝐺 ≅ ∠𝐾 G K G Prove: ∆𝐹𝐺𝐻 ~ ∆𝐽𝐾𝐿 G F H G J G L G ̅̅̅̅ 𝐹𝐺 ≅ ̅̅̅̅ 𝐺𝐻 Given ̅̅̅ 𝐽𝐾 ≅ ̅̅̅̅ 𝐾𝐿 Given ∠𝐹 ≅ ∠𝐽 Given ∆𝐹𝐺𝐻 ~ ∆𝐽𝐾𝐿 Angle-Angle Similarity Postulate ∆𝐹𝐺𝐻 is isosceles Definition of an isosceles triangle ∆𝐽𝐾𝐿 is isosceles Definition of an isosceles triangle ∠𝐹 ≅ ∠𝐻 Base angles of an isosceles triangle are congruent ∠𝐽 ≅ ∠𝐿 Base angles of an isosceles triangle are congruent ∠𝐻 ≅ ∠𝐽 ∠𝐻 ≅ ∠𝐿 Transitive Property of Congruency Transitive Property of Congruency ̅̅̅̅ 𝐹𝐺 ∥ ̅̅̅ 𝐾𝐽 Given ∠𝐺 𝑎𝑛𝑑 ∠𝐽 are alternate interior angles Definition of alternate interior angles ∠𝐺 ≅ ∠𝐽 If parallel lines are cut by a transversal, then alternate interior angles are congruent ∠𝐺𝐻𝐹 ≅ ∠𝐽𝐻𝐾 ∆𝐹𝐺𝐻 ~ ∆𝐾𝐽𝐻 Vertical angles are congruent Angle-Angle Similarity Postulate 𝐹𝐺 𝐺𝐻 = 𝐽𝐾 𝐾𝐿 Given ∠𝐺 ≅ ∠𝐾 Given Choose a point X on ̅̅̅ 𝐾𝐽 so that Through a point not on a line, there is exactly one line parallel to the KX = GF. Then draw given line ̅̅̅̅ ̅. 𝑋𝑌 ∥ 𝐽𝐿 ∠𝐾𝑋𝑌 ≅ ∠𝐽 If parallel lines are cut by a transversal, then corresponding angles are congruent ∆𝐾𝑋𝑌 ~ ∆𝐾𝐽𝐿 Angle-Angle Similarity Postulate 𝐾𝑋 𝑋𝑌 𝐾𝑌 = = 𝐾𝐽 𝐽𝐿 𝐾𝐿 Corresponding sides of similar triangles are proportional 𝐺𝐹 𝐺𝐻 = 𝐾𝐽 𝐾𝐿 Substitution 𝐾𝑋 𝐺𝐻 = 𝐾𝐽 𝐾𝐿 Substitution 𝐾𝑋 𝐾𝑌 𝐺𝐻 = = 𝐾𝐽 𝐾𝐿 𝐾𝐿 Substitution KY = GH Substitution in proportional relationship ∆𝐺𝐹𝐻 ≅ ∆𝐾𝑋𝑌 Side-Angle-Side Theorem of Triangle Congruency ∠𝐹 ≅ ∠𝐾𝑋𝑌 Corresponding parts of congruent triangles are congruent ∠𝐹 ≅ ∠𝐽 Transitive property of congruency ∆𝐺𝐹𝐻 ~ ∆𝐾𝐽𝐿 Angle-Angle Similarity Postulate
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