Question

ABCD is a trapezium in which AB||CD. If angle ADC = 2 * angle ABC, AD = a cm and CD = b cm, then the length (in cm) of AB is :-(A) (a/2) + 2b(B) a + b(C) (2a/b) + b(D) a + (2b/3)​

Answer 1

Answer:

Given that ABCD is a trapezium,

AB∥CD and AD=BC

To prove: (i) ∠A=∠B

(ii) ∠C=∠D

(iii) △ABC≅△BAD

(iv) AC=BD

Construction: Draw CE∥AD and extend AB to intersect CE at E.

Poof:

(i) As AECD is a parallelogram.

By construction, CE is parallel to AD and AE is parallel to CD,

∴AD=EC

But, AD=BC (Given)

∴BC=EC

⇒∠3=∠4 (Angle opposite to equal side are equal)

Now, ∠1+∠4=180

0

(consecutive interior angle of parallelogram )

And ∠2+∠3=180

0

(linear pair )

⇒∠1+∠4=∠2+∠3 ⇒∠1=∠2∵∠3=∠4 (so get cancelled with each other )

⇒∠A=∠B

(ii) AB∥CD

∴ ∠A+∠D=∠B+∠C=180

o

[Angles on the same side of transversal]

But, ∠A=∠B

Hence, ∠C=∠D

(iii) In △ABC and △BAD, we have

BC=AD (Given)

AB=BA (Common)

∠A=∠B (Proved)

Hence, by SAS congruence criterion,

△ABC≅△BAD

(iv) We have proved that, △ABC≅△BAD

∴ AC=BD (CPCT)

Hence, all the four required results have been proved.

I hope it will help you