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In a trapezium $ABCD, AB \parallel DC$ and its diagonals intersect at $O$. Prove that $\frac{OA}{OC} = \frac{OB}{OD}$·
In a trapezium $ABCD, AB \parallel DC$ and its diagonals intersect at $O$. Prove that $\frac{OA}{OC} = \frac{OB}{OD}$·
This Question has 1 answers.
### Given:
In trapezium $ABCD$, $AB \parallel DC$, and diagonals $AC$ and $BD$ intersect at $O$.
### To Prove:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $
### Proof:
Since $AB \parallel DC$, the diagonals $AC$ and $BD$ intersect each other at $O$. We apply the **Basic Proportionality Theorem (BPT)** or **Thales' theorem**, which states:
"If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio."
Consider $\triangle ADC$:
Since $AB \parallel DC$ and $O$ lies on diagonals $AC$ and $BD$, by the Basic Proportionality Theorem:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $
### Conclusion:
Hence, we have proved that:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $
In trapezium $ABCD$, $AB \parallel DC$, and diagonals $AC$ and $BD$ intersect at $O$.
### To Prove:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $
### Proof:
Since $AB \parallel DC$, the diagonals $AC$ and $BD$ intersect each other at $O$. We apply the **Basic Proportionality Theorem (BPT)** or **Thales' theorem**, which states:
"If a line is drawn parallel to one side of a triangle, it divides the other two sides in the same ratio."
Consider $\triangle ADC$:
Since $AB \parallel DC$ and $O$ lies on diagonals $AC$ and $BD$, by the Basic Proportionality Theorem:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $
### Conclusion:
Hence, we have proved that:
$ \displaystyle \frac{OA}{OC} = \frac{OB}{OD} $
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