Question

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Answer 1

Answer:

Hi friend!

Step-by-step explanation:

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,

Angle A +Angle E=180°

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

Angle E=180°-Angle A

Also ,BC =CE

Angle E=Angle CBE

[ABE is a straight line]

ANGLE ABC=180°-(180°-Angle A)

Angle ABC=180°-180°+Angle A

Angle B =Angle A.....(i)

(ii) Angle A +angle D =angle B+Angle C=180°

(angles on the same side of transversal)

Angle A +Angle D =Angle A +Angle C

(Angle A = Angle B) from eq (i)

Angle D =Angle C

(iii) in ∆ABC and ∆BAD,

AB=AB(Comman)

Angle DBA=Angle CBA(from eq (I)

AD=BC (given)

∆ABC =~∆BAD (by SAS Congurence rule )

(iv) Diagonal AC =Diagonal BD

(by CPCT as ∆ABC =~∆BAD

Hope this helps!

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