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09 Oct 2018, 21:41
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
66% (02:01) correct
34%
(02:12)
wrong
based on 92
sessions
History
Date
Time
Result
Not Attempted Yet
If A, B, C, D, and E are points in a plane such that line CD bisects ∠ACB and line CB bisects right angle ∠ACE, then ∠DCE =
(A) 22.5°
(B) 45°
(C) 57.5°
(D) 67.5°
(E) 72.5°
(A) 22.5°
(B) 45°
(C) 57.5°
(D) 67.5°
(E) 72.5°
10 Oct 2018, 11:02
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Bunuel
Angle ACE = 90 degree since CB bisects ACE then
BCE = 45 degree
ACB = 45 degree. Since CD bisects ACB
We get two angles ACD = 45/2 = 22.5 and DCB = 45/2 = 22.5
DCE = DCB + BCE = 22.5 + 45 = 67.5
Answer choice D
Attachments
Screen Shot 2018-10-10 at 9.11.21 PM.png [ 24.5 KiB | Viewed 6832 times ]
10 Oct 2018, 22:11
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Salsanousi
Great Image!! That is a perfect explanation.
