gottabwise wrote:

I totally messed up this problem. Don't mind my answer choice. I was rushing to get a MGMAT question bank completed. I've reworked the problem and understand it.

Why I was incorrect: I have in my head that a median from a vertex bisects the opposite side. I assumed AC bisected BD which is probably want the question maker wanted me to think. I didn't assume that the line just bisected the opposite side. I specifically remembered medians.

Can someone explain why my understanding of medians was incorrect? I want to learn from this mistake.

Thanks

Consider triangle ABD, angle ABD + angle ADB=90 (1)

Consider triangle ABC, angle ABD + angle BAD =90 (2)

Equating equations (1) and (2), since both of them equal to 90,

we get angle ADB= angle CAB

Two triangles can be similar if all the angles are equal. In this case, triangle ABC and triangle ABD are similar as:

angle ACB=angle BAD=90

angle ABD=angle ABC

angle ADB=angle CAB

The ratio of the sides opposites to the angles are equal. So,

\(\frac{AB(which is opposite to angle ADC)}{BC(which is opposite to angel CAB)}\) = \(\frac{BD(which is opposite to angle BAD)}{AB( which is opposite to angle ACB)}\)

Therefore,

\(\frac{5}{3}\)=\(\frac{3+x}{5}\)

On solving we get x=\(\frac{16}{3}\)