Question

ABCD is a trapezium in which AB is parallel to 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) triangle CEB is an isosceles triangle​

Answer 1

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Quadrilateral:

The closed figure formed by joining four non collinear points in an order is called a quadrilateral.

·Trapezium:

A quadrilateral in which one pair of opposite sides are parallel is called a trapezium.

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Given, ABCD is a trapezium in which AB||CD & AD=BC

To Show:

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

(iv) diagonal AC = diagonal BD

Construction: Draw a line through C parallel to DA intersecting AB produced at E.

Proof:

i)

AB||CD(given)

AD||EC (by construction)

So ,ADCE is a parallelogram

CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

We know that ,

∠A+∠E= 180°

[interior angles on the same side of the transversal AE]

∠E= 180° - ∠A

Also, BC = CE

∠E = ∠CBE= 180° -∠A

∠ABC= 180° - ∠CBE

[ABE is a straight line]

∠ABC= 180° - (180°-∠A)

∠ABC= 180° - 180°+∠A

∠B= ∠A………(i)

(ii) ∠A + ∠D = ∠B + ∠C = 180°

(Angles on the same side of transversal)

∠A + ∠D = ∠A + ∠C

(∠A = ∠B) from eq (i)

∠D = ∠C

(iii) In ΔABC and ΔBAD,

AB = AB (Common)

∠DBA = ∠CBA(from eq (i)

AD = BC (Given)

ΔABC ≅ ΔBAD

(by SAS congruence rule)

(iv) Diagonal AC = diagonal BD

(by CPCT as ΔABC ≅ ΔBAD)

Answer 2

Step-by-step explanation:

by sameer