Question

ABCD is a trapezium in which AB is parallel to CD and AD=BC show that
1, angle A=angle B
2,angle C=angle D
3, triangle ABC is congruent to triangle BAD
4, diagonal AC=diagonal BD

Answer 1

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Step-by-step explanation:

We have AB || CD and AD = BC

(i) To prove that ∠A = ∠B.

Produce AB to E and draw CE || AD.

∴

AB || DC

⇒ AE || DC

[Given]

Also

AD || CE

∴AECD is a parallelogram.

⇒AD = CE[opposite sides of the parallelogram AECD] ButAD = BC[Given] ∴BC = CE BC = CE

Now, in ΔBCE, we have

BC = CE ⇒∠CBE = ∠CEB...(1)

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

Also, ∠ABC + ∠CBE = 180°

[Linear pair] ...(2)

and ∠A + ∠CEB = 180°[∵Adjacent angles of a parallelogram are supplementary]

...(3)

From (2) and (3), we get

∠ABC + ∠CBE = ∠A + ∠CEB

But

∠CBE = ∠CEB

∴

∠ABC = ∠A

or

∠B = ∠A

or

∠A = ∠B

(ii) To prove that ∠C = ∠D.

AB || CD and AD is a transversal.

∠A + ∠D = 180°

[Sum of interior opposite angles]

Similarly, ∠B + ∠C = 180°

⇒∠A + ∠D = ∠B + ∠C

But ∠A = ∠B

[Proved]

∴∠C = ∠D

(iii) To prove ΔABC ≌ ΔBAD

In ΔABC and ΔBAD, we have

AB = BA

[Common]

BC = AD

[Given]

∠ABC = ∠BAD

[Proved]

∴ΔABC ≌ ΔBAD

[Using SAS criteria]

(iv) To prove that diagonal AC = diagonal BD

ΔABC ≌ ΔBAD

[Proved]

∴ Their corresponding parts are equal.

⇒the diagonal AC = the diagonal BD.