(vi) Δ ABC Δ DEF.ABCD is a trapezium in which AB II CAD+ BC (see Fig

1.	(vi) Δ ABC Δ DEF.ABCD is a trapezium in which AB II CAD+ BC (see Fig. 8.23). Show that(ii) KABCEA BAD(iv) diagonal AC diagonal BD
Answer» 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)

(vi) Δ ABC Δ DEF.ABCD is a trapezium in which AB II CAD+ BC (see Fig. 8.23). Show that(ii) KABCEA BAD(iv) diagonal AC diagonal BD

Answer»

Construction: Draw a line through C parallel to DA

intersecting AB produced at E.

Proof:

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)