In the given figure, \[AD=3cm,\,\,AE=5cm,\,\,BD=4cm,\,\,CE=4cm,\,\,CF=2cm,\,\,BF=2.5cm\], then find the pair of parallel lines and hence their lengths. |
Answer:
\[\frac{EC}{EA}=\frac{CF}{FB}\] and \[\frac{CF}{FB}=\frac{2}{2.5}=\frac{4}{5}\] \[\Rightarrow \frac{EC}{EA}=\frac{CF}{FB}\] In \[\Delta \text{ }ABC,\text{ }EF\parallel AB\] [Converse of Thales' theorem] Also, \[\frac{CE}{CA}=\frac{4}{4+5}=\frac{4}{9}\] \[\frac{CE}{CB}=\frac{2}{2+2.5}=\frac{2}{4.5}=\frac{4}{9}\] \[\frac{EC}{EA}=\frac{CF}{CB}\] \[\angle ECF=\angle ACB\] [Common] \[\Delta \text{ }CFE\tilde{\ }\Delta \text{ }CBA\] [SAS similarity] \[\Rightarrow \frac{EF}{AB}=\frac{CE}{CA}\] [In similar \[\Delta 's\], corresponding sides are proportional] \[\Rightarrow \frac{EF}{7}=\frac{4}{9}\] \[[\because ~\,AB=3+4=7\text{ }cm]\] \[\therefore EF=\frac{28}{9}cm\] and \[AB=7\,cm\]
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