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13 May 2019, 20:22
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Difficulty:
75%
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Question Stats:
44% (02:54) correct
56%
(02:47)
wrong
based on 25
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ABC is a triangle. D is a point on AB such that AD:DB=3:2 . E is the midpoint on BC. AE cuts CD at P. Find the ratio of CP:PD ?
A. 5:4
B. 2:1
C. 5:3
D. 4:3
E. 5:2
A. 5:4
B. 2:1
C. 5:3
D. 4:3
E. 5:2
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tri.png [ 7.47 KiB | Viewed 5377 times ]
14 May 2019, 10:35
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Hi, can anyone offer a solution to the problem?
14 May 2019, 21:22
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Draw a line, DF parallel to AE.
BDF is similar to BAE
AD/BD = FE/BF
3/2=FE/BF
if FE=3k, then BF=2k
E is the midpoint of BC, Hence EC= 3k+2k=5k
Also, CPE is similar to CDF
CP/PD = EC/FE
CP/PD= 5k/3k=5/3
BDF is similar to BAE
AD/BD = FE/BF
3/2=FE/BF
if FE=3k, then BF=2k
E is the midpoint of BC, Hence EC= 3k+2k=5k
Also, CPE is similar to CDF
CP/PD = EC/FE
CP/PD= 5k/3k=5/3
abhishek31
Attachments
tri.png [ 7.31 KiB | Viewed 5169 times ]
