Answered

The diagonals of a quadrilateral ABCD intersecting each other at the point O such that AO/BO=CO/DO.Show that ABCD is a trapizum

Answer :

sunshineallah9753

Answer:

Hi

✨ Good Morning ✨

Hope you are doing well and hope this helps you ☺️ ☺️ ☺️ ☺️

Given that the diagonals of quadrilateral ABCD intersect at point O such that \( \frac{AO}{BO} = \frac{CO}{DO} \), we need to show that ABCD is a trapezium.

Let's denote the given ratio by \( k \), i.e., \( \frac{AO}{BO} = \frac{CO}{DO} = k \).

### Proof:

1. **Consider the triangles formed by the diagonals and the point O:**

Since \( \frac{AO}{BO} = k \), by the properties of similar triangles (since triangles AOD and BOC share the same angle at O and have sides proportional due to the ratio \( k \)), we have:

\[

\triangle AOD \sim \triangle BOC

\]

2. **Using Menelaus' theorem in triangles AOB and COD:**

Menelaus' theorem states that for a transversal intersecting three collinear points (in this case, A, O, B and C, O, D respectively), the product of the ratios of the segments is equal to 1.

Applying Menelaus' theorem to triangle AOB with transversal COD gives us:

\[

\frac{AO}{BO} \cdot \frac{BC}{AC} \cdot \frac{OD}{OD} = 1

\]

Simplifying this gives:

\[

k \cdot \frac{BC}{AC} \cdot 1 = 1 \Rightarrow \frac{BC}{AC} = \frac{1}{k}

\]

Similarly, applying Menelaus' theorem to triangle COD with transversal AOB gives:

\[

\frac{CO}{DO} \cdot \frac{AB}{AD} \cdot \frac{BO}{AO} = 1

\]

Simplifying this gives:

\[

k \cdot \frac{AB}{AD} \cdot \frac{1}{k} = 1 \Rightarrow \frac{AB}{AD} = 1

\]

3. **Conclusion:**

From the above results, we have \( \frac{BC}{AC=1/k \) and \( \frac{AB}{AD} = 1 \).

For ABCD to be a trapezium, one pair of opposite sides must be parallel. From the ratios derived:

-BC/AC =1/K implies BC and AC are not equal unless \( k = 1 \), but generally, they are not.

- AB/AD = 1 implies AB = AD.

Therefore, AB and AD are parallel (since AB = AD), making ABCD a trapezium with AB and CD as its parallel sides.

Thus, we have shown that under the condition AO/BO= CO/DO , the quadrilateral ABCD must be a trapezium.

♥If you need further clarification feel free to ask ♥

✿Have a nice day ahead dear✿⁠

(⁠｡⁠♡⁠‿⁠♡⁠｡⁠)(⁠｡⁠♡⁠‿⁠♡⁠｡⁠)(⁠｡⁠♡⁠‿⁠♡⁠｡⁠)

Other Questions