In the adjoining figure, $A B C D$ is a trapezium in which $A B ∥ D C$ . The diagonals $A C$ and $B D$ intersect at $O$ . Prove that $A O / O C = B O / O D$ .

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Solution

From the given figure, $A B C D$ is a trapezium in which $A B ∥ D C$ ,
The diagonals $A C$ and $B D$ intersect at $O$ .
So we have to prove that, $A O / O C = B O / O D$
Consider the $△ A O B$ and $△ C O D$ ,
$∠ A O B = ∠ C O D$ … [vertically opposite angles]
$∠ O A B = ∠ O C D$
Therefore, $∠ A O B \sim △ C O D$
So, $O A / O C = O B / O D$
Now by using above result we have to find the value of $x$ if $O A = 3 x - 19, O B = x - 4, O C = x - 3$ and $O D = 4$ .
$O A / O C = O B / O D$
$(3 x - 19) / (x - 3) = (x - 4) / 4$
By cross multiplication we get,
$(x - 3) (x - 4) = 4 (3 x - 19)$
$X^{2} - 4 x - 3 x + 12 = 12 x - 76$
$X^{2} - 7 x + 12 - 12 x + 76 = 0$
$X^{2} - 19 x + 88 = 0$
$X^{2} - 8 x - 11 x + 88 = 0$
$X (x - 8) - 11 (x - 8) = 0$
$(x - 8) (x - 11) = 0$
Take $x - 8 = 0$
$X = 8$
Then, $x - 11 = 0$
$X = 11$
Therefore, the value of $x$ is $8$ and $11$ .

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