Question

117. ABCD is a trapezium in which AB II CD and AD=BC, show that
A
B.
T
D
С
(i) LA=/B
(ii) /C=D
(iii) Δ ABC αΔ BAD
(iv) Diagonal AC=Diagonal BD​

Answer 1

Answer:

I hope this will help you .

Answer 2

Question-

ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that

(i )∠A=∠B

(ii )∠C=∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD

Answer-

i) Produce AB to E and draw CF || AD.. .(1)

∵ AB || DC

⇒ AE || DC Also AD || CF

∴ AECD is a parallelogram.

⇒ AD = CE …(1)

[ ∵ Opposite sides of the parallelogram are equal]

But AD = BC …(2) [Given]

By (1) and (2), BC = CF

Now, in ∆BCF, we have BC = CF

⇒ ∠CEB = ∠CBE …(3)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° … (4)

[Linear pair]

and ∠A + ∠CEB = 180° …(5)

[Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From (3)]

⇒ ∠B = ∠A …(6)

(ii) AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180° …(7) [Co-interior angles]

Similarly, ∠B + ∠C = 180° … (8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]