question_answer

Prove ?If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio?.

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.