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F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
02 Aug 2017, 08:37
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19% (02:14) correct
80% (02:14) wrong based on 67 sessions
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#GREpracticequestion F is the midpoint of AE, and D is the midpoint of CE.jpg [ 20.98 KiB  Viewed 3264 times ]
F is the midpoint of AE, and D is the midpoint of CE. Which of the following statements MUST be true? Indicate all such statements. ❑ FD is parallel to AC. ❑ The area of triangle DEF equals the area of triangle CDF. ❑ The area of triangle ABF is less than the area of triangle DEF. ❑ Angle AFB equals angle BFC. ❑ The area of triangle ACE equals AC x BF.
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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
24 Sep 2017, 08:34
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Any hint for this question?



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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
25 Sep 2017, 02:34
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A is right Because DE is \(\frac{1}{2}\) of CE and FE is \(\frac{1}{2}\) of AE (i.e., corresponding sides have proportional lengths), and angle DEF is shared between /the two triangles, you can see that triangles ACE and FDE must be similar. Similar triangles have the further property that corresponding angles are of equal measure. Thus, for example, angle DFE equals angle CAE, and so FD is parallel to AC. B is right Looking at the two smaller triangles in the right half of the figure, you can see that triangle CD F and triangle DEF have collinear and equal “bases” (CD and DE), and share their third vertex (F), which is some fixed distance away from CE. Because CE is comprised of bases CD and DE of triangles CD F and DEF, respectively, the two triangles have the same height. Because CDF and DEF have equal bases and the same height, they must have the same area. (For a similar reason, triangles ACF and FCE must have equal areas; more on that later.) Choice E is right Because triangles ACF and FCE must have equal areas as indicated above, you can see that the area of triangle ACE must be twice that of triangle ACF. Note that FD is parallel to AC due to (true) choice (A), and BF is perpendicular toFD. Therefore, BF must be perpendicular to AC as well. Put differently, AC can be regarded as the base, and BF the height, of triangle ACF. The area of triangle ACF equals ^ times AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF.
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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
27 Jun 2018, 08:54
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Hello guys, any hint why choices C and D are false?



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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
28 Jun 2018, 02:21
if you redraw the figure as follow, then C and D not necessarily are true. Attachment:
triangle.png [ 35.83 KiB  Viewed 4099 times ]
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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
08 Jan 2019, 02:57
thanks



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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
04 Aug 2019, 15:17
For answer choice E, how do you know triangles ACF and FCE are equal in area ? Thanks in advance!



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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
04 Aug 2019, 16:37
Because CE is comprised of bases CD and DE of triangles CD F and DEF, respectively, the two triangles have the same height. Because CDF and DEF have equal bases and the same height, they must have the same area. (For a similar reason, triangles ACF and FCE must have equal areas; more on that later.) Read carefully. Ask if you do need further explanations. Sir Regards
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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh [#permalink]
09 Dec 2019, 05:17
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Carcass wrote: A is right
Because DE is \(\frac{1}{2}\) of CE and FE is \(\frac{1}{2}\) of AE (i.e., corresponding sides have proportional lengths), and angle DEF is shared between /the two triangles, you can see that triangles ACE and FDE must be similar. Similar triangles have the further property that corresponding angles are of equal measure. Thus, for example, angle DFE equals angle CAE, and so FD is parallel to AC.
B is right
Looking at the two smaller triangles in the right half of the figure, you can see that triangle CD F and triangle DEF have collinear and equal “bases” (CD and DE), and share their third vertex (F), which is some fixed distance away from CE. Because CE is comprised of bases CD and DE of triangles CD F and DEF, respectively, the two triangles have the same height. Because CDF and DEF have equal bases and the same height, they must have the same area. (For a similar reason, triangles ACF and FCE must have equal areas; more on that later.)
Choice E is right
Because triangles ACF and FCE must have equal areas as indicated above, you can see that the area of triangle ACE must be twice that of triangle ACF. Note that FD is parallel to AC due to (true) choice (A), and BF is perpendicular toFD. Therefore, BF must be perpendicular to AC as well. Put differently, AC can be regarded as the base, and BF the height, of triangle ACF. The area of triangle ACF equals ^ times AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF. Choice E is right
Because triangles ACF and FCE must have equal areas as indicated above, you can see that the area of triangle ACE must be twice that of triangle ACF. Note that FD is parallel to AC due to (true) choice (A), and BF is perpendicular toFD. Therefore, BF must be perpendicular to AC as well. Put differently, AC can be regarded as the base, and BF the height, of triangle ACF. The area of triangle ACF equals ^ times AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF.Edited : (The last line will be) The area of triangle ACF equals = \(\frac{1}{2}\) * AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF. Extended: ACF = \(\frac{1}{2}\) * AC x BF => 2 *ACF = AC x BF so, ACE = AC x BF (as we know triangle ACE is twice of triangle ACF)
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Re: F is the midpoint of AE, and D is the midpoint of CE. Wh
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09 Dec 2019, 05:17





