In isosceles trapezoid ABCD, AD = BC. what is angle a + angle c?
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Through B, draw a straight line parallel to AD which meets CD at E.
Now, since AB∥DE and AD∥BE,
∴ABED is a parallelogram.
Thus, ED=AB=18cm
As, BE∥AD and CD is a transversal,
∴∠BED=∠D=60°
(∵∠ABE=∠D)
Since, in an isosceles trapezium, the base angles are equal, ∠C=∠D=60°
In ΔBEC,∠BEC+∠ECD+∠CBE=180°
(Angle sum property of a triangle)
⇒60° +60° +∠CBE=180°
⇒120° +∠CBE=180°
⇒∠CBE=180° −120°
⇒∠CBE=60°
As the measure of ∠BEC=60°,∠ECB=60° and ∠CBE=60° ,ΔCBE is an equilateral triangle, Thus
∴CE=BC=12cm(BC=AD=12cm)
Now, CE+ED=12+18
⇒DC=30cm
Hope this'll help you.
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Answer:
isoceles triangles are those whose any two sides are equal and similar to other triangle then the angle between these two sides will also be equal
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