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 392 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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