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question of class 9 from NCERT book Chapter 8
Answers
Answer:
Solution:
Given: In trapezium ABCD, AB || CD and AD = BC.
To Prove: (i) ∠A = ∠B (ii) ∠C = ∠D
(iii) ΔABC ≅ ΔBAD (iv) diagonal AC = diagonal BD
Constructions: Join AC and BD. Extend AB and draw a line through C parallel to DA meeting AB produced at E.
Proof:
(i) Since AB || DC ⇒ AE || DC and AD || CE [by construction]
Now, since opposite pairs of sides are parallel
⇒ ADCE is a parallelogram
⇒ AD = CE …(v) [Opposite sides of a ||gm]
But AD = BC [Given]
⇒ BC = CE
Now in ΔBCE, BC = CE
⇒ ∠E = ∠CBE [Angles opposite equal sides]
Also, ∠A + ∠E = 180° [Co-interior angles] …(i)
∠B + ∠CBE = 180° [Linear pair]
∴ ∠B + ∠E = 180° [Putting ∠E = ∠CBE] …(ii)
From (i) and (ii) we have
∠A + ∠E = ∠B + ∠E
⇒ ∠A = ∠B
(ii) ∠A + ∠D = 180° [Co-interior angles]
∠B + ∠C = 180° [Co-interior angles]
⇒ ∠A + ∠D = ∠B + ∠C
⇒ ∠D = ∠C [∵ ∠A = ∠B]
⇒ ∠C = ∠D
(iii) In ΔABC and ΔBAD, we have
AD = BC [Given]
∠A = ∠B [Proven]
AB = CD [Common]
⇒ ΔABC ≅ ΔBAD [ASA congruence]
(iv) diagonal AC = diagonal BD [by CPCT]
Step-by-step explanation:
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