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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
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Hi PathFinder007,

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.

Answer B: 1/100 = 1/200 + 1/200...TRUE. Keep it.

Answer C: 1/100 = 0...NOT true...ELIMINATE it.

Answer D: 100 = 800...NOT true. ELIMINATE it.

Answer E: 100 = 0...NOT true. ELIMINATE it.

Final Answer:

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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
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PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.


Here is how we can arrive at the relation from scratch:

In right triangle ADB, \(BD^2 = AB^2 - AD^2\)

We need BD and AB in our equation but not AD. So get rid of AD.

Triangle ABC is similar to ADB. So \(\frac{AB}{AD} = \frac{BC}{BD}\)

\(AD = \frac{AB*BD}{BC}\)

Substitute in equation above:

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

\(BD^2*BC^2 = AB^2 * BC^2 - AB^2 * BD^2\)

Divide by \(AB^2 * BC^2 * BD^2\)

\(\frac{1}{AB^2} = \frac{1}{BD^2} - \frac{1}{BC^2}\)

\(\frac{1}{BD^2} = \frac{1}{AB^2} + \frac{1}{BC^2}\)

Answer (B)
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
VeritasPrepKarishma wrote:
PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.


Here is how we can arrive at the relation from scratch:

In right triangle ADB, \(BD^2 = AB^2 - AD^2\)

We need BD and AB in our equation but not AD. So get rid of AD.

Triangle ABC is similar to ADB. So \(\frac{AB}{AD} = \frac{BC}{BD}\)

\(AD = \frac{AB*BD}{BC}\)

Substitute in equation above:

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

\(BD^2*BC^2 = AB^2 * BC^2 - AB^2 * BD^2\)

Divide by \(AB^2 * BC^2 * BD^2\)

\(\frac{1}{AB^2} = \frac{1}{BD^2} - \frac{1}{BC^2}\)

\(\frac{1}{BD^2} = \frac{1}{AB^2} + \frac{1}{BC^2}\)

Answer (B)


Hi Karishma,
How do you deduce that "Triangle ABC is similar to ADB."
Kindly help...
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
ani781 wrote:
VeritasPrepKarishma wrote:
PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.


Here is how we can arrive at the relation from scratch:

In right triangle ADB, \(BD^2 = AB^2 - AD^2\)

We need BD and AB in our equation but not AD. So get rid of AD.

Triangle ABC is similar to ADB. So \(\frac{AB}{AD} = \frac{BC}{BD}\)

\(AD = \frac{AB*BD}{BC}\)

Substitute in equation above:

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

\(BD^2*BC^2 = AB^2 * BC^2 - AB^2 * BD^2\)

Divide by \(AB^2 * BC^2 * BD^2\)

\(\frac{1}{AB^2} = \frac{1}{BD^2} - \frac{1}{BC^2}\)

\(\frac{1}{BD^2} = \frac{1}{AB^2} + \frac{1}{BC^2}\)

Answer (B)


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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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
Expert Reply
ani781 wrote:
VeritasPrepKarishma wrote:
PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.


Here is how we can arrive at the relation from scratch:

In right triangle ADB, \(BD^2 = AB^2 - AD^2\)

We need BD and AB in our equation but not AD. So get rid of AD.

Triangle ABC is similar to ADB. So \(\frac{AB}{AD} = \frac{BC}{BD}\)

\(AD = \frac{AB*BD}{BC}\)

Substitute in equation above:

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

\(BD^2*BC^2 = AB^2 * BC^2 - AB^2 * BD^2\)

Divide by \(AB^2 * BC^2 * BD^2\)

\(\frac{1}{AB^2} = \frac{1}{BD^2} - \frac{1}{BC^2}\)

\(\frac{1}{BD^2} = \frac{1}{AB^2} + \frac{1}{BC^2}\)

Answer (B)


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.

Here are a couple of posts that will help you understand how to quickly identify similar triangles:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... -trianges/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03 ... -the-gmat/
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
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:

\(AC^2 = AB^2 + BC^2\) ...(2)

Now let's combine equations (1) and (2)

First, square equation 1: \(AB^2*BC^2 = BD^2*AC^2\)

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:

\(\frac{1}{BD^2} = \frac{AB^2}{AB^2*BC^2} + \frac{BC^2}{AB^2*BC^2}\)

\(\frac{1}{BD^2} = \frac{1}{BC^2}+ \frac{1}{AB^2}\) ...And this now matches one of our answer choices!

Answer B
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.


Let angle BAD = 37
angle BCA = 53
angle CBD = 37
and angle ABD is 53

sin37 = 3/5 and cos 37= 4/5
applying in triangle
BD=3 , AB =5 , BC =15/4

putting the values in expression B

1/ (bd)^2 = 1/9. LHS
RHS is 1/BC^2 + 1?BD^2 = (4/25)^2 + (1/5)^2 = 16/225 + 1/25 = (16+9) /225 = 25/225 = 1/9 = LHS

Ans is B :thumbup:
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
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PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.



Attachment:
Triangle.PNG


VERITAS PREP OFFICIAL SOLUTION:

What makes this question particularly difficult is the abstraction it entails; no numbers are provided in either the stimulus or the answer choices. But you do know multiple things about these triangles that should help you make sense of them:

1) Pythagorean Theorem \((a^2 + b^2 = c^2)\), which means that:

\((AB)^2 + (BC)^2 = (AC)^2\)

\((BD)^2 + (DC)^2 = (BC)^2\)

and

\((BD)^2 + (AD)^2 = (AB)^2\)


2) Area of a triangle equals \(\frac{1}{2}(base)(height)\), so since the area could be taken with AB as the height and BC as the base *OR* with BD as the height and AC as the base, you have:

\((AB)(BC) = (BD)(AC)\)

Now, using the Substitution Method for systems of equations you could look to get all the terms down to just AB, BD, and BC as you see in the answer choices, you could also pick numbers that will work with Pythagorean Theorem and check the answer choices. Since this is a right triangle but not necessarily isosceles, try the first set of Pythagorean triplets you can find, which is likely the 3:4:5 triangle. If you say that AB = 3, BC = 4, and AC = 5, you can then also say that BD needs to be 12/5, since the area calculations have to match with (AB)(BC) = (AC)(BD). Then if you plug in for the answer choices, you'll see that squaring (12/5) will lead to a messy fraction while squaring the other terms leads to integers, so none of A, D, and E can be correct. Then try B: \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}+\frac{1}{(AB)^2}\) would be:

\(\frac{25}{(144)}=\frac{1}{16}+\frac{1}{9}\)

Finding a common denominator on the right, you'll see that B is correct, as:

as:

\(\frac{25}{(144)}=\frac{9}{144}+\frac{16}{144}\).
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
Hi Experts / Members, in what scenario will the perpendicular from angle B divide the hypotenuse into two equal halves (AD = DC)?

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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
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VIVA1060 wrote:
Hi Experts / Members, in what scenario will the perpendicular from angle B divide the hypotenuse into two equal halves (AD = DC)?

chetan2u Bunuel


It would happen when BA = BC, so when a right triangle is an isosceles. In this case the perpendicular to the hypotenuse would be the median and the angle bisector.
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink]
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