Question

. ABCD is a quadrilateral. If AD = BC
and <BAD =<ABC, prove that ABCD
is an isosceles trapezium.​

Answer 1

Hi friend!

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!