Question

in a Trapezium ABCD in which ab is parallel to CD then answer is angle A + Angle B is equals to straight line angle , angle b + angle c is equals to straight line angle​

Answer 1

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)

!!!Hope this helps!!!

❤️BRAINLIEST PLS ❤️