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BD is parallel to AE. [#permalink]
02 Aug 2017, 08:31
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BD is parallel to AE.
Quantity A 
Quantity B 
xz 
wy 
A) Quantity A is greater. B) Quantity B is greater. C) The two quantities are equal. D) The relationship cannot be determined from the information given.
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Re: BD is parallel to AE. [#permalink]
23 Sep 2017, 18:54
Explain Please



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Re: BD is parallel to AE. [#permalink]
23 Sep 2017, 20:23
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Pria wrote: Explain Please Here given BD is parallel to AE, so angle CBD = angle CAE, and angle BDC = angle AEC. Therefore the triangles BCD and ACE are similar (Angle C is common to both triangles and by AAA triangle BCD and triangle ACE are similar) Now it is given side BC = x , AB = y and AC =x+w. side CD = y, DE = z and CE = y+Z Now as both triangles BCD and ACE are similar therfore we have \(\frac{CB}{CA}\) = \(\frac{CD}{CE}\) substitute the values we get \(\frac{x}{(x+w)}\)= \(\frac{y}{(y+z)}\) or x(y+z) = y(x+w) or xy + xz = xy + wy or xz = wy. So option C.
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Re: BD is parallel to AE. [#permalink]
24 Sep 2017, 07:36
Thank you.



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Re: BD is parallel to AE. [#permalink]
26 Sep 2017, 17:04
I don't understand. CA could be different length than CE
Can anyone explain? Thank you



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Re: BD is parallel to AE. [#permalink]
26 Sep 2017, 20:48
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pclawong wrote: I don't understand. CA could be different length than CE
Can anyone explain? Thank you \(\frac{{CB}}{{CA}} = \frac{{CD}}{{CE}}\) because they are similar and the ratio's of there length has to be equal. CA= x+w (total length of CA, x & w are given) CE = y+z (total length of CE, y & z are given)
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Re: BD is parallel to AE. [#permalink]
12 Nov 2018, 18:59
pranab01 wrote: Pria wrote: Explain Please Here given BD is parallel to AE, so angle CBD = angle CAE, and angle BDC = angle AEC. Therefore the triangles BCD and ACE are similar (Angle C is common to both triangles and by AAA triangle BCD and triangle ACE are similar) Now it is given side BC = x , AB = y and AC =x+w. side CD = y, DE = z and CE = y+Z Now as both triangles BCD and ACE are similar therfore we have \(\frac{CB}{CA}\) = \(\frac{CD}{CE}\)substitute the values we get \(\frac{x}{(x+w)}\)= \(\frac{y}{(y+z)}\) or x(y+z) = y(x+w) or xy + xz = xy + wy or xz = wy. So option C. If you put the reason, it will be better If two triangles are similar, the ratio of their corresponding sides are equal.  Please let me know, if I am wrong.



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Re: BD is parallel to AE. [#permalink]
09 Jan 2019, 12:41
thank you



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Re: BD is parallel to AE. [#permalink]
11 Jan 2019, 14:53
pclawong wrote: I don't understand. CA could be different length than CE
Can anyone explain? Thank you By maintaining ratio, any length is possible.



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Re: BD is parallel to AE. [#permalink]
09 May 2019, 04:02
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A line drawn parallel to a side of the triangle divides the two other sides proportionally. So x/w and y/z have the same ratio. Hence xz and wy are equal



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Re: BD is parallel to AE. [#permalink]
16 May 2019, 04:02
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Hi,
This is a question of proportionality of sides of two triangles that are similar. The larger and the smaller triangles are similar due to two common angles.
Here, x/w=y/z
So, xz=wy.
Hence, option (C) is the right answer.



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Re: BD is parallel to AE. [#permalink]
30 Sep 2019, 06:16
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Let's say x = a*w where 'a' is the constant term of proportionality between a side of the little triangle and the corresponding side of the larger triangle. Then, it must be also that: y = a*z. Therefore: x*z = (a*w)*z w*y = w*(a*z) The two are equal, hence C.
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Re: BD is parallel to AE. [#permalink]
03 May 2020, 10:13
grewhiz wrote: Hi,
This is a question of proportionality of sides of two triangles that are similar. The larger and the smaller triangles are similar due to two common angles.
Here, x/w=y/z
So, xz=wy.
Hence, option (C) is the right answer. Can we reliably predict this relation? I thought the only relation we can predict from the diagram is x/(x+w) = y/(y+z)




Re: BD is parallel to AE.
[#permalink]
03 May 2020, 10:13





