In the given figure, if AD = BC and AD || BC, then prove that AB = DC. [2 Marks]

Open in App

Solution

In $△ A C B$ and $△ C A D$ ,
$A D = B C$ [Given]
$∠ C A D = ∠ A C B$ [Alternate angles] [1 Mark]

$C A = A C$ [common side]
$△ A C B ≅ △ C A D$ [SAS congruency]
$∴ A B = D C$ [CPCT] [1 Mark]

Suggest Corrections

Similar questions

In the given figure, if AD = BC and AD || BC, then prove that AB = DC.

∠ D A B = ∠ D C B .

In figure, $∠ B A C = 90^{\circ}$ and segment $A D ⊥ B C$ . Prove that ${A D}^{2} = B D \times D C$ . [3 MARKS]

{A D}^{2}

\times

Join BYJU'S Learning Program