Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

视频信息

答案文本 复制

视频字幕 复制

We have a trapezium ABCD where AB is parallel to DC. The diagonals AC and BD intersect at point O. We need to prove that AO divided by OC equals OB divided by OD. To solve this problem, we will identify two triangles formed by the intersection. Consider triangle AOB and triangle COD. We need to prove these triangles are similar using angle criteria. Since AB is parallel to DC, we can identify equal angles. Angle BAO equals angle DCO as alternate interior angles. Similarly, angle ABO equals angle CDO as alternate interior angles. Also, angle AOB equals angle COD as vertically opposite angles. By the AA similarity criterion, triangle AOB is similar to triangle COD. For similar triangles, the ratios of corresponding sides are equal. Therefore, AO over CO equals BO over DO, which gives us our required result: AO over OC equals OB over OD. To summarize what we have learned: We proved that triangles AOB and COD are similar using parallel line properties and the AA similarity criterion. This established the required ratio relationship for any trapezium with intersecting diagonals.