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Since this question asks which of the following relationships MUST be true, we can TEST VALUES to find the true statement and disprove the others. Here's how.

Since we have 2 "small" right triangles inside of 1 "big" right triangle, I'm going to take advantage of some common right triangle patterns that exist in math.

Let's make AngleA = 45 degrees and AngleC = 45 degrees

So both little triangles are 45/45/90 and the big triangle is 45/45/90

This means that AD = BD = CD and AB = BC

If I make AD = BD = CD = 10... .....then AB = BC = 10(root2)...

Now we can TEST those values against the answers and eliminate...

Answer A: 100 = 200 + 200...NOT true...ELIMINATE it.

Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]

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13 Jan 2016, 20:01

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Hi Karishma, How do you deduce that "Triangle ABC is similar to ADB." Kindly help...

Triangles ADB and ABC are similar by Angle-Angle (AA) similarity condition. Angle (A) is common to both the triangles as well as there is 1 right angle (=90 degree) in both the triangles. Thus you get the similarity of the triangles.
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Hi Karishma, How do you deduce that "Triangle ABC is similar to ADB." Kindly help...

In the given figure, triangle ABC is similar to triangle ADB and triangle ABC is also similar to triangle BDC. Such a figure should instantly make you think of similarity. The reason of similarity is the AA property. Each of these triangles has a right angle so one set of angles is equal. Then, triangles ABC and ADB have angle A common so two angles are equal. Hence triangles ABC and ADB are similar. Also triangles ABC and BDC have angle C common so two angles are equal. Hence triangles ABC and BDC are similar.

Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]

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15 Jan 2016, 09:11

I love problems like this because there are so many different approaches.

I must say though, I don't like Bunuel's approach, since he starts with the right answer, then shows that it is the right answer. But how did he know to start there in the first place?!

Let's work out the answer starting with what we know about the triangles.

Looking at \(\triangle\)ABC, we can express the area in two different ways:

\(Area = \frac{1}{2}AB*BC = \frac{1}{2}BD*AC\)

So \(AB*BC = BD*AC\) ...(1)

Since AC doesn't appear in the answer choices, we'll have to convert that into something else. Use Pythagoras:

Then replace \(AC^2\) using equation (2): \(AB^2*BC^2 = BD^2*(AB^2 + BC^2)\)

Ok, now we have an expression involving only what we want, AB, BC and BD. We need to try to manipulate it into one of the answer choices. All the answer choices have BD on one side and AB and BC on the other, so we can start by doing that:

\(BD^2 = \frac{AB^2*BC^2}{(AB^2 + BC^2)}\)

This equation looks like a form we should be familiar with (from all those work rate problems). If we flip the equation over (take the reciprocal of both sides) it will look like:

\(\frac{1}{BD^2} = \frac{(AB^2 + BC^2)}{AB^2*BC^2}\) Now this is something we've seen many times. The equation can be rewritten as:

Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]

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17 Jun 2017, 23:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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