Difference between revisions of "AA similarity"

(Created page with "Theorem: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar. Proof: Let ABC and DEF be two triangles such that <math>\angle ...")
 
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Proof:
 
Proof:
 
Let ABC and DEF be two triangles such that <math>\angle A = \angle D</math> and <math>\angle B = \angle E</math>.
 
Let ABC and DEF be two triangles such that <math>\angle A = \angle D</math> and <math>\angle B = \angle E</math>.
<math>\angle A + \angle B + \angle C = 180</math> (Sum of all angles in a triangle is <math>180</math>)
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<math>\angle A + \angle B + \angle C = 180</math> and
<math>\angle D + \angle E + \angle F = 180</math> (Sum of all angles in a triangle is <math>180</math>)
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<math>\angle D + \angle E + \angle F = 180</math>  
<math>\angle A  + \angle B + \angle C=\angle D + \angle E + \angle F</math>
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Thus, we can write the equation: <math>\angle A  + \angle B + \angle C=\angle D + \angle E + \angle F \Longrightarrow
<math>\angle D + \angle E + \angle C = \angle D + \angle E + \angle F</math> (since <math>\angle A = \angle D</math> and <math>\angle B = \angle E</math>)
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\angle D + \angle E + \angle C = \angle D + \angle E + \angle F</math>, since we know that <math>\angle A = \angle D</math> and <math>\angle B = \angle E</math>, from before.
<math>\angle C = \angle F</math>.
+
Therefore, by subtracting <math>\angle D + \angle E</math> by both equations, we get <math>\angle C = \angle F</math>.

Revision as of 22:47, 24 January 2016

Theorem: In two triangles, if two pairs of corresponding angles are congruent, then the triangles are similar.

Proof: Let ABC and DEF be two triangles such that $\angle A = \angle D$ and $\angle B = \angle E$. $\angle A + \angle B + \angle C = 180$ and $\angle D + \angle E + \angle F = 180$ Thus, we can write the equation: $\angle A  + \angle B + \angle C=\angle D + \angle E + \angle F \Longrightarrow \angle D + \angle E + \angle C = \angle D + \angle E + \angle F$, since we know that $\angle A = \angle D$ and $\angle B = \angle E$, from before. Therefore, by subtracting $\angle D + \angle E$ by both equations, we get $\angle C = \angle F$.