Question

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that
(i)∠A = ∠B
(ii)∠C = ∠D
(iii)ΔABC ≅ ΔBAD
(iv)diagonal AC = diagonal BD

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]​

Answer 1

$\huge\mathbb{SOLUTION:-}$

To Construct:

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

(i)CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

, BC = CE

⇒ ∠CBE = ∠CEB

also,

∠A + ∠CBE = 180° (Angles on the same side of transversal and ∠CBE = ∠CEB)

∠B + ∠CBE = 180° ( As Linear pair)

⇒ ∠A = ∠B

(ii)∠A + ∠D = ∠B + ∠C = 180° (Angles on the same side of transversal)

⇒ ∠A + ∠D = ∠A + ∠C (∠A = ∠B)

⇒ ∠D = ∠C

(iii)In ΔABC and ΔBAD,

AB = AB (Common)

∠DBA = ∠CBA

AD = BC (Given)

, ΔABC ≅ ΔBAD [SAS congruency]

(iv)Diagonal AC = diagonal BD by CPCT as ΔABC ≅ ΔBA.

Answer 2

$\large{\sf{\boxed{\boxed{Answer}}}}$

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)