If a straight line is drawn parallel to one side of a triangle, then it divides the other two sides proportionally.
Data: In \[ \Delta ABC\]
DE II BC
To prove: \[\frac{AD}{DB}=\frac{AE}{EC}\]
Construction: 1. Join DC and BE. 2. Draw EL ꓕ AB and DN ꓕ AC.
Proof:
| STATEMENT |
REASON |
| \[\frac{Area of \Delta ADE}{Area of \Delta BDE}=\frac{\frac{1}{2}\times AD\times EL}{\frac{1}{2}\times DB\times EL}\] |
\[(\because A=\frac{1}{2}\times b\times h)\] |
| \[\therefore \frac{ \Delta ADE}{ \Delta BDE}=\frac{AD}{DB}\] |
|
| \[\frac{ \Delta ADE}{ \Delta CDE}=\frac{\frac{1}{2}\times AD\times DN}{\frac{1}{2}\times EC\times DN}\] |
\[(\because A=\frac{1}{2}\times b\times h)\] |
| \[\therefore \frac{ \Delta ADE}{ \Delta CDE}=\frac{AE}{EC}\] |
|
| \[\Rightarrow \frac{ AD}{ DB}=\frac{AE}{CE}\] |
[\[\because \Delta BDE= \Delta CDE \] and Axiom-1] |
Axiom1: For two triangles, if two pairs of sides have the same ratio then those two triangles are similar.
