Question

ABCD is a trapezium in which AB | DC and AD = BC. If CE is drawn parallel
to AD meeting AB at E, prove the following:
(a) AECD is a parallelogram
(b) AD = EC
(c) ACEB is an isosceles triangle
A
106/Quadrilaterals
E​

Answer 1

Answer:

thankyou for

Step-by-step explanation:

please correct part c

Answer 2

Answer:

Given: ABCD is a trapezium where AB∣∣CD and AD=BC

Construction : Extends AB and draw a line through C point to DA intersecting AB produced at E

Prrof: AD∣∣CE (from construction) &

AE∣∣DC (AS AB∣∣CD, & AB is extended)

AECD is a parallelogram.

In AECD, both pair of opposite sides are parallel.

∴AD=CE (opposite sides of parallelogram are equal)

But AD=BC (Given)

⇒BC=CE

So, ∠CEB=∠CBE ...(1) ( In ΔBCE, angles opposite to equal sides are equal)

For AD∣∣CE,

& AE is the transversal,

∠A+∠CEB=180

o

[interior angles on same side of transversal is supplementary]

∠A=180

o

−∠CEB ....(2)

Also AE is line,

so, ∠B+∠CE=180

o

(liner pairs)

∠B+∠CBE=180

0

(from (1))

∠B=180

o

−∠CBE ....(3)

from (2) and (3)

∠A=∠B

∴ The answer is ∠A=∠B

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