$A C$ and $B D$ are the diagonals of a quadrilateral $A B C D$ , prove that: $\to A B + \to D C = \to A C + \to D B$ .

Open in App

Solution

Let $A B C D$ be a rectangle, $A C$ and $B D$ its diagonals.
In $Δ A B C$ ,
$\to A B + \to B C = \to A C$
$\to A B = \to A C - \to B C$ ...........(i)
In $Δ B C D$ by vector law of addition
$\to D C = \to D B + \to B C$ ......(ii)
by adding (i) and (ii), we get
$\to A B + \to D C = \to A C - \to B C + \to D B + \to B C$
$\to A B + \to D C = \to A C + \to D B$

Suggest Corrections

Similar questions

M

N

A C

B D

A B C D

\to A B + \to A D + \to C B + \to C D = 4 \to M N

A B C D E

\to A B + \to A E + \to B C + \to D C + \to E D + \to A C

\to A B + \to A D + \to C B + \to C D = 4 \to E F

Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

A C

B D

A B C D

O

a r (△ A O D) = a r (△ B O C)

A B C D

Join BYJU'S Learning Program