Question

ABCD is an isosceles trapezium with angleB = 60° and AD ||BC. Side AD is produced till E such that AE = BC, CE is joined. Find angleDCE.

Plz answer only if you know, plzzzzz

Answer 1

Answer:

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=60o(∵∠ABE=∠D).

Since, in an isosceles trapezium, the base angles are equal, ∠C=∠D=60o.

In ΔBEC,∠BEC+∠ECD+∠CBE=180o

(Angle sum property of a triangle)

⇒60o+60o+∠CBE=180o

⇒120o+∠CBE=180o

⇒∠CBE=180o−120o

⇒∠CBE=60o

As the measure of ∠BEC=60o,∠ECB=60o and ∠CBE=60o,ΔCBE is an equilateral triangle, Thus

∴CE=BC=12cm(BC=AD=12cm)

Now, CE+ED=12+18

⇒DC=30cm