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.

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## Comments

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TopNewestLet 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.

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

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I uploaded the image here

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