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In the figure above, AB is perpendicular to BC and BD is perpendicular

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

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18 Dec 2014, 09:45
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In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.
[Reveal] Spoiler: OA

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Triangle.PNG [ 18.29 KiB | Viewed 28074 times ]

Last edited by Bunuel on 18 Dec 2014, 10:16, edited 1 time in total.
Renamed the topic and edited the question.

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

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18 Dec 2014, 10:29
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Option B: $$\frac{1}{(BD)^2} = \frac{1}{(BC)^2} + \frac{1}{(AB)^2}$$;

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

Since (AB)^2 + (BC)^2 = (AC)^2, then $$\frac{1}{(BD)^2} = \frac{(AC)^2}{(BC)^2*(AB)^2}$$;

$$(BC)^2*(AB)^2 = (AC)^2*(BD)^2$$;

$$(BC*AB)^2 = (AC*BD)^2$$.

The area of the triangle is 1/2*BC*AB as well as 1/2*AC*BD. Thus BC*AB = AC*BD.

Therefore $$(BC*AB)^2 = (AC*BD)^2$$ is true.

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

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20 Dec 2014, 19:48
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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.

[Reveal] Spoiler:
B

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Special Offer: Save $75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3423 [2], given: 172 SVP Status: The Best Or Nothing Joined: 27 Dec 2012 Posts: 1853 Kudos [?]: 2633 [7], given: 193 Location: India Concentration: General Management, Technology WE: Information Technology (Computer Software) Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink] Show Tags 30 Dec 2014, 02:56 7 This post received KUDOS 1 This post was BOOKMARKED Refer diagram below: Attachment: Triangle.PNG [ 4.01 KiB | Viewed 20992 times ] Looking at the OA, its clear that we require to draw relation between BD, AB & BC (All coloured) Say AB = 3 & BC = 4, so AC = 5 & BD would be $$= \frac{12}{5}$$(As area of triangle $$\frac{1}{2} AB*BC = \frac{1}{2} AC*BD$$) Checking the OA Option A $$(\frac{12}{5})^2 \neq{3^2 + 4^2}$$ Option B $$(\frac{5}{12})^2 = \frac{1}{4^2} + \frac{1}{3^2}$$ Answer = B _________________ Kindly press "+1 Kudos" to appreciate Kudos [?]: 2633 [7], given: 193 GMAT Club Legend Joined: 09 Sep 2013 Posts: 16535 Kudos [?]: 274 [0], given: 0 Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink] Show Tags 13 Jan 2016, 20:01 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. _________________ Kudos [?]: 274 [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7677 Kudos [?]: 17405 [0], given: 232 Location: Pune, India Re: In the figure above, AB is perpendicular to BC and BD is perpendicular [#permalink] Show Tags 14 Jan 2016, 02:51 Expert's post 1 This post was BOOKMARKED 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) _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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

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14 Jan 2016, 14:52
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 $$AB^2 * BC^2 * BD^2$$

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

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

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]

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14 Jan 2016, 15:13
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 $$AB^2 * BC^2 * BD^2$$

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

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

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]

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14 Jan 2016, 21:11
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 $$AB^2 * BC^2 * BD^2$$

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

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

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:
http://www.veritasprep.com/blog/2014/03 ... -trianges/
http://www.veritasprep.com/blog/2014/03 ... -the-gmat/
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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:

$$AC^2 = AB^2 + BC^2$$ ...(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!

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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).

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

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28 Aug 2017, 08:47
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.

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
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Re: In the figure above, AB is perpendicular to BC and BD is perpendicular   [#permalink] 28 Aug 2017, 08:47
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