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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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Question Stats:

80% (00:46) correct 19% (00:12) wrong based on 31 sessions

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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.

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Please let me know, if I am wrong.
Re: BD is parallel to AE.   [#permalink] 12 Nov 2018, 18:59
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