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# BD is parallel to AE.

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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.
[Reveal] Spoiler: OA

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Re: BD is parallel to AE. [#permalink]  23 Sep 2017, 18:54
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Re: BD is parallel to AE. [#permalink]  23 Sep 2017, 20:23
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Pria wrote:

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:

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
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