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Similar Triangles: Why isn't AC a median?

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Manager
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Similar Triangles: Why isn't AC a median? [#permalink]

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19 Jan 2010, 21:03
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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
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Re: Similar Triangles: Why isn't AC a median? [#permalink]

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19 Jan 2010, 22:53
Please dont get confused with the medians..they have not stated anything about mid points.

This question involved similar triangles.

Consider 2 Concepts.

1..area of triangle ABD = 1/2 AB AD = 1/2 AC BD
here AB = 5 Pythagoras theore and AC = 4 given

2.. Triangles ABC and ABD are similar

Divide equation A by B

we get CD/BD = 16/25
now CD/BD = CD/CD+3 = 16/25

This given CD = 16/3
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Kudos [?]: 1911 [0], given: 235

Manager
Joined: 24 Jul 2009
Posts: 192

Kudos [?]: 47 [0], given: 10

Location: Anchorage, AK
Schools: Mellon, USC, MIT, UCLA, NSCU
Re: Similar Triangles: Why isn't AC a median? [#permalink]

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23 Jan 2010, 20:42
gurpreetsingh wrote:
Please dont get confused with the medians..they have not stated anything about mid points.

This question involved similar triangles.

Consider 2 Concepts.

1..area of triangle ABD = 1/2 AB AD = 1/2 AC BD
here AB = 5 Pythagoras theore and AC = 4 given

2.. Triangles ABC and ABD are similar

Divide equation A by B

we get CD/BD = 16/25
now CD/BD = CD/CD+3 = 16/25

This given CD = 16/3

Thanks for the explanation. I understand part one but am totally confused by part two, specifically AC/AB=CD/AD.
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Re: Similar Triangles: Why isn't AC a median? [#permalink]

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23 Jan 2010, 23:32
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 ABD=angle ABC

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=$$\frac{16}{3}$$

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Manager
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Re: Similar Triangles: Why isn't AC a median? [#permalink]

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09 Feb 2010, 19:35
this is confusing...

how are you refering to angle ABC when you are talking about triangle ABD ?
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Manager
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Re: Similar Triangles: Why isn't AC a median? [#permalink]

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09 Feb 2010, 20:52
another approach

AC=4 BC =3, ABC is right traingle so ab=5
let cd = x, so ad^2=16+x^2
25 + 16 + x^2 = (3+x)^2
solving x=16/3

D

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Re: Similar Triangles: Why isn't AC a median?   [#permalink] 09 Feb 2010, 20:52
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