fskilnik wrote:
In the figure shown, AB=AC and CE=CF. What is the value of x?
(A) 90
(B) 110
(C) 120
(D) 130
(E) 140
Source:
http://www.GMATH.net (Exercise 12 of the
Quant Class #02 included in our free test drive!)
Attachment:
GMATH_figure028EDIT (2).jpg [ 19.86 KiB | Viewed 507 times ]
We need only four properties:
• Angles opposite congruent sides are congruent
• Vertical angles are congruent
• Angles that are colinear sum to 180°
• The sum of interior angles of a triangle = 180°
Given: ∠ F = 40°
∠ F = ∠ CEF = 40° (angles opposite congruent sides are congruent)
Sum of the two congruent angles
(40° + 40°) = 80°
In ∆ CEF, the third angle,
∠ECF must = (180°- 80°) =
100°∠ ECF and ∠ ACB, are colinear
∠ ACB must = (180° - 100°) = 80°
∠ACB = ∠ABC = 80° (opposite sides AB = AC)
Vertical angles: ∠CEF = ∠BED
From above
∠CEF = 40° = ∠BEDThe two known angles of ∆BDE sum to (40° + 80°) = 120°
Third angle BDE = (180° - 120°) = 60°
Finally, because the angles are colinear:
\((x+∠BDE) = 180°\)
\((x+60°)=180°\)\(x=120°\)Answer C
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