Name: Trapezium and Its Properties
Uploaded: 2019-07-19
Duration: 3 min 54 s
Description: A trapezium is a special convex quadrilateral having only one pair of opposite sides parallel. This video explains the diagrammatic representation of a trape...

Question

A B C D

is a parallelogram. The bisector of

∠ A

and

∠ B

meet at

E

, which lies on

D C

. Prove that

A D = \frac{1}{2} A B

.

Accepted Answer

$A B C D$ is a parallelogram, so we know
$A B ∥ C D$ and $B C ∥ D A$

Here $∠ D A E = ∠ E A B$ (As given $A E$ is angle bisector of $∠ B A D$ )
And
$∠ C B E = ∠ E B A$ ( As given $B E$ is angle bisector of $∠ A B C$ )
Now we take $A E$ as transversal line and we know $A B ∥ C D$
So
$∠ E A B = ∠ D E A$ (As they are alternate interior angles)
So,
$∠ D E A = ∠ D A E$ (As we know $∠ D A E = ∠ E A B$ )
So,
$A D = D E$ (As we know base angle theorem, if base angles of any triangle is equal than opposite sides are also equal)

And
Now we take $A E$ as transversal line and we know $A B ∥ C D$
So
$∠ E B A = ∠ C E B$ (As they are alternate interior angles)
So,
$∠ C B E = ∠ C E B$ (As we know $∠ C B E = ∠ E B A$ )
So,
$B C = E C$ (As we know base angle theorem , if base angles of any triangle is equal than opposite sides are also equal )
Hence
$A D = D E = B C = E C . . . . . (i)$
We know
$A B = C D$
And
$A B = D E + E C (As C D = D E + E C)$
$A B = A D + A D$ from equation $(i)$
$A D = \frac{1}{2} A B$ Hence proved.

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