Could someone please solve this?
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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
o
(∵∠ABE=∠D).
Since, in an isosceles trapezium, the base angles are equal, ∠C=∠D=60
o
.
In ΔBEC,∠BEC+∠ECD+∠CBE=180
o
(Angle sum property of a triangle)
⇒60
o
+60
o
+∠CBE=180
o
⇒120
o
+∠CBE=180
o
⇒∠CBE=180
o
−120
o
⇒∠CBE=60
o
As the measure of ∠BEC=60
o
,∠ECB=60
o
and ∠CBE=60
o
,ΔCBE is an equilateral triangle, Thus
∴CE=BC=12cm(BC=AD=12cm)
Now, CE+ED=12+18
⇒DC=30cm
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