Quadrilateral is A' B' C' D' is a dilation of quadrilateral ABCD about point P with a scale factor of 3.
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Answer:
m∠CEB is 55°
Step-by-step explanation:
Since ∠ADE = 55°, and ∠ADE is half of ∠ADC because ED bisects ∠ADC. Bisect means to cut in half.
∠ADC = 110° because it is double of ∠ADE.
Since AB║CD and AD║BC, the two sets of parallel lines means this shape is a parallelogram. In parallelograms, <u>opposite angles have equal measures</u>.
∠ADC = ∠CBE = 110°
All quadrilaterals have a sum of angles 360°. Since ∠DCB = ∠BAD and we know two of these other angles are each 110°:
360° - 2(110°) = 2(∠DCB)
∠DCB = 140°/2
∠DCB = ∠BAD = 70°
∠DCB was bisected by EC, which makes each divided part half.
∠DCE = ∠BCE = (1/2)(∠DCB)
∠DCE = ∠BCE = (1/2)(70°)
∠DCE = ∠BCE = 35°
All triangles' angles sum to 180°.
In ΔBCE, ∠BCE = 35° and ∠CBE = 110°.
∠CEB = 180° - (∠BCE + ∠CBE)
∠CEB = 180° - (35° + 110°)
∠CEB = 55°
Therefore m∠CEB is 55°.
