In $\triangle A B C, A C=A B=25$ and $B C=40 . D$ is a point chosen on $B C$. From $D$, perpendiculars are drawn to meet $A C$ at $E$ and $A B$ at $F . D E+D F$ equals
(A) 12
(B) 35
(C) 24
(D) 25
(E) $\frac{35}{2} \sqrt{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-144.jpg?height=276&width=464&top_left_y=388&top_left_x=1275)
In $\triangle A B C, A C=A B=25$ and $B C=40 . D$ is a point chosen on $B C$. From $D$, perpendiculars are drawn to meet $A C$ at $E$ and $A B$ at $F . D E+D F$ equals
(A) 12
(B) 35
(C) 24
(D) 25
(E) $\frac{35}{2} \sqrt{2}$

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-292.jpg?height=322&width=547&top_left_y=213&top_left_x=1407)

## Solution

We start by drawing a line from $A$ that is perpendicular to the base $C B$. Since $\triangle A B C$ is isosceles, $M$ is the midpoint of $C B$ thus making $C M=M B=20$. Using pythagoras in $\triangle A C M$ we find $A M$ to be $\sqrt{25^{2}-20^{2}}=15$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-292.jpg?height=314&width=543&top_left_y=683&top_left_x=1409)

Join $A$ to $D$. The area of $\triangle A B C$ is $\frac{1}{2}(40)(15)=300$ but it is also, $\frac{1}{2}(E D)(25)+\frac{1}{2}(D F)(25)$

$$
=\frac{25}{2}(E D+D F)
$$

Therefore, $E D+D F=\frac{2}{25}(300)=24$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-292.jpg?height=306&width=539&top_left_y=1056&top_left_x=1411)

ANSWER: (C)