Answer:
Given, In \[\Delta \text{ }ABC,\text{ }DE\parallel BC\] To prove: \[\frac{AD}{DB}=\frac{AE}{EC}\] Construction: Draw \[EM\bot AB\] and \[DN\bot AC\]. Join B to E and C to D. Proof: In \[\Delta \text{ }ADE\] and \[\Delta \text{ }BDE\] \[\frac{ar(\Delta \,ADE)}{ar(\Delta \,BDE)}=\frac{\frac{1}{2}\times AD\times EM}{\frac{1}{2}\times DB\times EM}=\frac{AD}{DB}\] ?(i) [Area of \[\Delta =\frac{1}{2}\times \] base \[\times \] corresponding altitude] In \[\Delta \text{ }ADE\] and \[\Delta \text{ }CDE\] \[\frac{ar(\Delta \,ADE)}{ar(\Delta \,CDE)}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times EC\times DN}=\frac{AE}{EC}\] ?(ii) Since, \[DE\parallel BC\] [Given] \[\therefore ar(\Delta \,BDE)=ar(\Delta \,CDE)\] ...(iii) [\[\Delta s\] on the same base and between the same parallel sides are equal in area] From eq. (i), (ii) and (iii) \[\frac{AD}{DB}=\frac{AE}{EC}\] Hence Proved.
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