# Difference between revisions of "2013 AMC 12A Problems/Problem 9"

Epicwisdom (talk | contribs) (Created page with "Note that because <math>\bar{DE}</math> and <math>\bar{EF}</math> are parallel to the sides of <math>\triangle ABC</math>, the internal triangles <math>\triangle BDE</math> and <...") |
Epicwisdom (talk | contribs) m (Replaced bar with overline in LaTeX) |
||

Line 1: | Line 1: | ||

− | Note that because <math>\ | + | Note that because <math>\overline{DE}</math> and <math>\overline{EF}</math> are parallel to the sides of <math>\triangle ABC</math>, the internal triangles <math>\triangle BDE</math> and <math>\triangle EFC</math> are similar to <math>\triangle ABC</math>, and are therefore also isosceles triangles. |

It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>. | It follows that <math>BD = DE</math>. Thus, <math>AD + DE = AD + DB = AB = 28</math>. | ||

Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>. | Since opposite sides of parallelograms are equal, the perimeter is <math>2 * (AD + DE) = 56</math>. |

## Revision as of 04:26, 7 February 2013

Note that because and are parallel to the sides of , the internal triangles and are similar to , and are therefore also isosceles triangles.

It follows that . Thus, .

Since opposite sides of parallelograms are equal, the perimeter is .