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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
1 year, 9 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.

Maria Kozlowska - 1 year, 9 months ago

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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

Greg Grapsas - 1 year, 9 months ago

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I uploaded the image here

Maria Kozlowska - 1 year, 9 months ago

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@Maria Kozlowska 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

Greg Grapsas - 1 year, 9 months ago

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