Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a simi

Question

Diagonals $AC$ and $BD$ of a trapezium $ABCD$ with $AB \parallel DC$ intersect each other at the point $O$. Using a similarity criterion for triangles, show that $\frac{AO}{OC} = \frac{BO}{OD}$.

Accepted Answer

Solution: Given: A trapezium $ABCD$ in which $AB \parallel DC$ and diagonals $AC$ and $BD$ intersect at point $O$. To Show: $\frac{AO}{OC} = \frac{BO}{OD}$ Proof Steps: Consider triangle $AOB$ and triangle $COD$. Since $AB \parallel DC$ and $BD$ is a transversal line, the alternate interior angles are equal. Therefore, $\angle ABO = \angle CDO$. Similarly, considering $AC$ as a transversal line between the parallel lines $AB$ and $DC$, the alternate interior angles are equal. Therefore, $\angle BAO = \angle DCO$. Also, the vertically opposite angles formed at the intersection point $O$ are equal. Therefore, $\angle AOB = \angle COD$. By using the AAA (Angle-Angle-Angle) similarity criterion, since all three corresponding angles of triangle $AOB$ are equal to the corresponding angles of triangle $COD$, we can write: $$	riangle AOB \sim 	riangle COD$$ We know that if two triangles are similar, the ratios of their corresponding sides are equal (proportional). Therefore, from the similarity of these triangles, we get: $$\frac{AO}{CO} = \frac{BO}{DO} = \frac{AB}{CD}$$ Taking the first two parts of the proportion, we have: $$\frac{AO}{OC} = \frac{BO}{OD}$| Hence, it is proved that $\frac{AO}{OC} = \frac{BO}{OD}$.

Triangles

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