# Solve for equal sides of isosceles triangle given base and two perpendiculars from base to the two sides

I cannot find the specific problem but it summarized as follows:

Isosceles triangle $ABC$ with $AB=AC$ and $BC=60$.
A point $D$ on base $BC$ is located such that a perpendicular to side $AB$, denoted as $DE$, has length 16 and a perpendicular to side $AC$, denoted as $DF$, has length 32. Points $E$ and $F$ are on $AB$ and $AC$ respectively. What is the length of the two equal sides of triangle $ABC$?

The correct answer shown was 50. I got a different answer using similar triangles etc. I checked and rechecked and continued to get my original answer. I then constructed the posted answer graphically and while the one perpendicular was equal to 16, it resulted in the other perpendicular being ~38.75.

In my answer, each of the two equal sides were ~43.148. Moreover, when I graphed the problem, the two perpendiculars measured 16 and 32 respectively.

At this point, I am wondering if my rendering of the image is wrong as no drawing was included in the problem.

Note by Greg Grapsas
4 years, 4 months ago

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Let point $G$ be foot of perpendicular from A on $BC$. $\triangle EBD$ is similar to $\triangle DFC$ and to $\triangle BGA$. $BD=20, BE=12, AG=40, AB=50$

It looks like the answer is in fact 50.

- 4 years, 4 months ago

Can you include a drawing? If point D is on the base then BE cant be 12 sonce it the hypotenuse of triangle EDB

- 4 years, 4 months ago

- 4 years, 4 months ago

Thank you...when i saw it i realized my stupid mistake...i had df correct but i foolishly had DE perpendicular to the base instead of BA and just couldnt see it until you sent me the drawing

- 4 years, 4 months ago