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 $ADEF$?
+2666 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 \(\overline{AD} + \overline{DE} + \overline{EF} + \overline{AF} = x + y + x + y = 2(x+y) = 2 \times 25 = \color{brown}\boxed{50}\)
+2666 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 \(\overline{AD} + \overline{DE} + \overline{EF} + \overline{AF} = x + y + x + y = 2(x+y) = 2 \times 25 = \color{brown}\boxed{50}\)
