Question

QS: ABCD is an isosceles trapezium where AB II CD and AD = BC. If DMand
CN are perpendiculars from D and C respectively, prove thatA ADM is
congruent to A BCN.
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Answer 1

Explanation:

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