QUESTION IMAGE
Question
in the figure, \\( \overline { d e } \parallel \overline { b c } \\) based on the figure, indicate whether each proportion is true or false. \\( \frac { a b } { b c } = \frac { a e } { e d } \\) \\( \frac { a d } { b a } = \frac { a e } { c a } \\) \\( \frac { a d } { a e } = \frac { a b } { a c } \\) \\( \frac { a d } { d b } = \frac { a e } { a c } \\) \\( \frac { d b } { d c } = \frac { b a } { c a } \\)
Step1: Use the basic proportionality theorem (Thales' theorem)
If \(DE\parallel BC\) in \(\triangle ABC\), then \(\frac{AD}{DB}=\frac{AE}{EC}\), \(\frac{AD}{AB}=\frac{AE}{AC}\), \(\frac{DB}{AB}=\frac{EC}{AC}\)
Step2: Analyze \(\frac{AB}{BC}=\frac{AE}{ED}\)
By Thales' theorem, there is no relation of the form \(\frac{AB}{BC}=\frac{AE}{ED}\) as the theorem relates the segments of the two sides \(AB\) and \(AC\) cut by the parallel line \(DE\). So, \(\frac{AB}{BC}=\frac{AE}{ED}\) is False.
Step3: Analyze \(\frac{AD}{BA}=\frac{AE}{CA}\)
Since \(\frac{AD}{AB}=\frac{AE}{AC}\) (by Thales' theorem), cross - multiplying gives \(AD\times AC = AE\times AB\). Also, \(\frac{AD}{BA}=\frac{AE}{CA}\) can be rewritten as \(AD\times CA=AE\times BA\). So, \(\frac{AD}{BA}=\frac{AE}{CA}\) is True.
Step4: Analyze \(\frac{AD}{AE}=\frac{AB}{AC}\)
From \(\frac{AD}{AB}=\frac{AE}{AC}\), cross - multiplying gives \(AD\times AC = AE\times AB\), and \(\frac{AD}{AE}=\frac{AB}{AC}\) also gives \(AD\times AC=AE\times AB\) by cross - multiplication. So, \(\frac{AD}{AE}=\frac{AB}{AC}\) is True.
Step5: Analyze \(\frac{AD}{DB}=\frac{AE}{AC}\)
By Thales' theorem \(\frac{AD}{DB}=\frac{AE}{EC}\), not \(\frac{AE}{AC}\). So, \(\frac{AD}{DB}=\frac{AE}{AC}\) is False.
Step6: Analyze \(\frac{DB}{DC}=\frac{BA}{CA}\)
By Thales' theorem \(\frac{AD}{DB}=\frac{AE}{EC}\), then \(\frac{DB}{AD}=\frac{EC}{AE}\), and \(\frac{DB + AD}{AD}=\frac{EC + AE}{AE}\) (by componendo). \(\frac{AB}{AD}=\frac{AC}{AE}\) or \(\frac{AB}{AC}=\frac{AD}{AE}\). Also, \(\frac{DB}{DC}
eq\frac{BA}{CA}\). So, \(\frac{DB}{DC}=\frac{BA}{CA}\) is False.
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| True | False | |
|---|---|---|
| \(\frac{AD}{BA}=\frac{AE}{CA}\) | \( \bigcirc\) | |
| \(\frac{AD}{AE}=\frac{AB}{AC}\) | \( \bigcirc\) | |
| \(\frac{AD}{DB}=\frac{AE}{AC}\) | \( \bigcirc\) | |
| \(\frac{DB}{DC}=\frac{BA}{CA}\) | \( \bigcirc\) |