the problem in your appraoch is that u have assumed wrong angles to be similar
to check if cd=3 then bd=6
and since ac=4 and bc=3 ab and ad will be 5
hence
5^2+5^2=6^2 which is not possible
now
the triangle similar are DCA,ACB, DCA
from tri DAB and ACB
ad/4=5/3
ad=20/3
now in tri DAB and DAC
cd/ad=4/5
-->cd=4/5*ad
=4/5*20/3=16/3

hence D
OA??
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This is a MGMAT Question,
OA is D and OE is as below. Hope it helps.

Because angles BAD and ACD are right angles, the figure above is composed of three similar right triangles: BAD, ACD and BCA. [Any time a height is dropped from the right angle vertex of a right triangle to the opposite side of that right triangle, the three triangles that result have the same 3 angle measures. This means that they are similar triangles. See below for further explanation.]

To solve for the length of side CD, we can set up a proportion, based on the relationship between the similar triangles ACD and BCA:

BC/CA = CA/CD

3/4 = 4/CD

CD = 16/3
The correct answer is D.

Addendum: Let's look at how we know that triangles ACD and BCA are similar.

1) Let's say that <CDA is x degrees, and <DAC is y degrees. Since <ACD is 90 degrees, and the sum of all the interior angles in a triangle is 180, we know that x + y = 90.

2) Now let's look at <BAC. We know that <BAC + <DAC = 90, since <BAD is labeled as a right angle. We also know that <DAC is y degrees (from step 1), and that x + y = 90. Putting these facts together, we know that <BAC is x degrees.

3) We know <ACB is a right angle, since <ACD is a right angle. Since <ACB is a right angle, <BAC + <CBA = 90. Given that <BAC is x degrees, <CBA must be y degrees.

4) To summarize, <CAB has the same measure as <CDA (x degrees) , and <CBA has the same measure as <DAC (y degrees). This means that in similar triangles CAB and CAD, side BC of CAB corresponds to side CA of CAD, and side CA of CAB corresponds to side CD of CAD.

Thus, BC/CA = CA/CD.

Again, the correct answer is D.

all three triangles abc, acd & abd are similar.
so, \frac{4}{3} = \frac{cd}{4} ... cd = \frac{16}{3}

it can't be \frac{4}{3} = \frac{4}{cd} ... cd = 3, because in that case 5^2+5^2 = 6^2 is not true
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