The figure shows square ABCD, point E on line BC, and point F on line CD. Moreover BE

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Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

by Max@Math Revolution » Tue Jun 09, 2020 4:15 am

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[GMAT math practice question]

The figure shows square ABCD, point E on line BC, and point F on line CD. Moreover BE = CF and ∠AEC = 110. What is ∠CBF?

6.9PS.png

A. 20
B. 30
C. 33
D. 40
E. 43

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i256: Junior | Next Rank: 30 Posts; Posts: 11; Joined: Sat May 30, 2020 5:24 am

Re: The figure shows square ABCD, point E on line BC, and point F on line CD. Moreover BE

by i256 » Tue Jun 09, 2020 9:48 am

Triangles AEB and BCF are similar triangles, 2 common sides 1 common angles (SAS congruency)

So, angle EBF = BAE = (110 - 90 ) = 20

Max@Math Revolution: Elite Legendary Member; Posts: 3991; Joined: Fri Jul 24, 2015 2:28 amFollowed by: 37 members; Location: Las Vegas, USA; Thanked: 19 times

Re: The figure shows square ABCD, point E on line BC, and point F on line CD. Moreover BE

by Max@Math Revolution » Thu Jun 11, 2020 12:31 am

=>

Since AB = BC, BE = CF and ∠ABE = ∠BCF = 90, triangles ABE and BCF are congruent.
Then we have ∠BFC = ∠AEB = 180° – 110° = 70°.
We have ∠CBF = 180° – (70° + 90°) = 20°.

Therefore, the answer is A.
Answer: A

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