12. ABCD is a trapezium in which AB || CD and AB=BC(see figure 8.23).show that:-
(i) Angle A= Angle B
(ii) Angle C = Angle D
(iii) Triangle ABC = Triangle BAD
(iv) diagonal AC = diagonal BD
[Hint : Extend AB and draw a line through C
parallel to DA intersecting AB produced at E.)
Answers
Step-by-step explanation:
Quadrilateral:
The closed figure formed by joining four non collinear points in an order is called a quadrilateral.
·Trapezium:
A quadrilateral in which one pair of opposite sides are parallel is called a trapezium.
=========================================================
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)
Answer:
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.
Step-by-step explanation: