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In $\triangle ABC,$ $AB=AC=25$ and $BC=23.$ Points $D,E,$ and $F$ are on sides $\overline{AB},$ $\overline{BC},$ and $\overline{AC},$ respectively, such that $\overline{DE}$ and $\overline{EF}$ are parallel to $\overline{AC}$ and $\overline{AB},$ respectively. What is the perimeter of parallelogram $ADEF$?

Aug 7, 2022

#1
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Let $$\overline{AD} = x$$ and $$\overline{BD} = y$$. We know that $$\overline{AB} = \overline{AD} + \overline{BD}$$ so $$x + y = 25$$

Because $$\triangle BDE$$ is similar to $$\triangle ABC$$, we know that $$\triangle BDE$$ is isoceles, meaning $$\overline{DE} = y$$

Also, because $$ADEF$$ is a parallelogram, we know that $$\overline{EF} = x$$ and $$\overline{AF} = y$$

So, the perimeter of the rectangle is just $$\overline{AD} + \overline{DE} + \overline{EF} + \overline{AF} = x + y + x + y = 2(x+y) = 2 \times 25 = \color{brown}\boxed{50}$$

Aug 7, 2022
edited by BuilderBoi  Aug 7, 2022

#1
+2541
0

Let $$\overline{AD} = x$$ and $$\overline{BD} = y$$. We know that $$\overline{AB} = \overline{AD} + \overline{BD}$$ so $$x + y = 25$$

Because $$\triangle BDE$$ is similar to $$\triangle ABC$$, we know that $$\triangle BDE$$ is isoceles, meaning $$\overline{DE} = y$$

Also, because $$ADEF$$ is a parallelogram, we know that $$\overline{EF} = x$$ and $$\overline{AF} = y$$

So, the perimeter of the rectangle is just $$\overline{AD} + \overline{DE} + \overline{EF} + \overline{AF} = x + y + x + y = 2(x+y) = 2 \times 25 = \color{brown}\boxed{50}$$

BuilderBoi Aug 7, 2022
edited by BuilderBoi  Aug 7, 2022