Since two corresponding pairs of angles are equal, OAB is similar to OCD by the AA (Angle-Angle) similarity criterion.

Since the ratio of corresponding sides of similar triangles is equal, we have (OA)/(OC) = (OB)/(OD). This completes the proof.

30. Diagonals $AC$ and $BD$ of trapezium $ABCD$ with $AB \parallel DC$ intersect each other at point $O$ . Show that $\frac{OA}{OC} = \frac{OB}{OD}$ . $\textit{\text{Diagram details for} 30}: A$ trapezium $ABCD$ with horizontal base $AB$ at the bottom, parallel top side $DC$ , and diagonals $AC$ and $BD$ intersecting at point $O$ . 31. $P$ and $Q$ are respectively the points on the sides $AB$ and $AC$ of a triangle $ABC$ such that $AB = 12.5\text{ cm}$ , $AP = 5\text{ cm}$ and $CQ = 6\text{ cm}$ . If $PQ \parallel BC$ , then find the length of $AQ$ . 32. In the given figure, $XZ$ is parallel to $BC$ . $AZ = 3\text{ cm}$ , $ZC = 2\text{ cm}$ , $BM = 3\text{ cm}$ , and $MC = 5\text{ cm}$ . Find the length of $XY$ . $\textit{\text{Diagram details for} 32}: A$ triangle $ABC$ with vertex $A$ at the top and base $BC$ at the bottom. Point $M$ lies on $BC$ . A line segment $XZ$ is drawn parallel to $BC$ (indicated by arrows on $XZ$ and $BC$ ), intersecting side $AB$ at $X$ and side $AC$ at $Z$ . A line segment is drawn from vertex $A$ to point $M$ on $BC$ , intersecting $XZ$ at point $Y$ . Given lengths are $AZ = 3\text{ cm}$ , $ZC = 2\text{ cm}$ , $BM = 3\text{ cm}$ , and $MC = 5\text{ cm}$ . 33. $E$ and $F$ are points on the sides $PQ$ and $PR$ respectively of a $\Delta PQR$ . If $PE = 3.9\text{ cm}$ , $EQ = 3\text{ cm}$ , $PF = 3.6\text{ cm}$ and $PR = 6\text{ cm}$ , find whether $EF \parallel QR$ .

Answer: Hence proved, $\frac{OA}{OC} = \frac{OB}{OD}$ .

Step-by-step solution

Step 1: Identify parallel lines and transversal pairs

In trapezium $ABCD$ , the sides $AB$ and $DC$ are parallel. The diagonals $AC$ and $BD$ act as transversals intersecting these two parallel sides.

Step 2: Establish angle equalities in triangles OAB and OCD

Considering transversal $AC$ , alternate interior angles are equal: $\angle OAB = \angle OCD$ . Considering transversal $BD$ , alternate interior angles are equal: $\angle OBA = \angle ODC$ . Additionally, $\angle AOB = \angle COD$ as vertically opposite angles.

Step 3: Apply AA similarity criterion

Since two corresponding pairs of angles are equal, $\triangle OAB$ is similar to $\triangle OCD$ by the AA (Angle-Angle) similarity criterion.

Step 4: Equate corresponding side ratios

Since the ratio of corresponding sides of similar triangles is equal, we have $\frac{OA}{OC} = \frac{OB}{OD}$ . This completes the proof.

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