Question

AD = BC and AD||BC is AB = DC give reason in support of your answer​.​

Answer 1

Given :

• AD = BC
• AD ∥ BC

To Prove :

• AB = DC

Proof :

We are given :

• AD = BC
• AD ∥ BC

As, opposite sides of Quadilateral are equal and parallel, Therefore it is a parallelogram.

Now, we know that opposite sides of parallelogram are equal and parallel.

Hence,

• AB = DC
• AB ∥ DC

Hence, we have proved that AB = DC.

Some basic properties of Parallelogram :

• The opposite sides of parallelogram are parallel and equal.
• The opposite angles of parallelogram are equal.
• The consecutive angles are supplementary to each other i.e. are equal to 180°.

Answer 2

Answer:

$\setlength{\unitlength}{1.4cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(8.6,3){\large\sf{D}}\put(7.7,0.9){\large\sf{A}}\put(11.1,0.9){\large\sf{B}}\put(8,1){\line(1,0){3}}\qbezier(11,1)(11.5,2)(12,3)\put(9,3){\line(3,0){3}}\multiput(8.4,2)(3,0){2}{\line(1,0){.2}}\qbezier(8,1)(8.5,2)(9,3)\qbezier(8,1)(10,2)(12,3)\put(12.1,3){\large\sf{C}}\end{picture}$

• AD = BC
• AD || BC
• Prove AB = DC

$\underline{\bigstar\:\textsf{In Triangles ADC and ABC :}}$

$:\implies\sf AD=BC\qquad(Given)\\\\\\:\implies\sf \angle ADC=\angle ABC\qquad(Alternate\: Interior\:Angles)\\\\\\:\implies\sf AC=AC\qquad(Common)\\\\\\:\implies\sf \Delta \:ADC \cong \Delta \:ABC\qquad(By\: SAS\: theorem)$

⠀

$\sf Now\ as\ \Delta \:ADC \cong \Delta \:ABC\ ,\ hence\\\\\dashrightarrow\sf AB = DC\qquad(By\ CPCT)$