Question

p304
ex. 4

1. if \\(\overline{ab} \parallel \overline{cd}\\), and \\(\overline{ac}\\) and \\(\overline{bd}\\) intersect at \\(e\\), prove that \\(de \times ae = be \times ce\\).
2. in \\(\triangle rst\\), angle \\(s\\) is a right angle and \\(\overline{ed} \perp \overline{rt}\\). prove that \\(dr \times ts = ed \times sr\\).

Explanation:

Ex.4

Step1: Use the property of similar triangles

Since \(AB\parallel CD\), \(\triangle ABE\sim\triangle CDE\) (by the AA similarity criterion: \(\angle AEB=\angle CED\) (vertically - opposite angles) and \(\angle BAE = \angle DCE\) (alternate interior angles as \(AB\parallel CD\))).
For similar triangles \(\triangle ABE\) and \(\triangle CDE\), we have the proportion \(\frac{AE}{CE}=\frac{BE}{DE}\).

Step2: Cross - multiply the proportion

Cross - multiplying the proportion \(\frac{AE}{CE}=\frac{BE}{DE}\) gives \(DE\times AE=BE\times CE\).

Step1: Use the property of similar triangles

In \(\triangle RST\), \(\angle S = 90^{\circ}\) and \(ED\perp RT\). So, \(\triangle RDE\sim\triangle RST\) (by the AA similarity criterion: \(\angle R=\angle R\) (common angle) and \(\angle RED=\angle S = 90^{\circ}\)).
For similar triangles \(\triangle RDE\) and \(\triangle RST\), we have the proportion \(\frac{DR}{SR}=\frac{ED}{TS}\).

Step2: Cross - multiply the proportion

Cross - multiplying the proportion \(\frac{DR}{SR}=\frac{ED}{TS}\) gives \(DR\times TS=ED\times SR\).

Answer:

\(DE\times AE = BE\times CE\) is proved.

Ex.5