Theorem:
"The areas of similar triangles are proportional to the squares on the corresponding sides".
Data: \[ \Delta ABC \approx \Delta DEF\]
\[\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\]
To prove: \[\frac{Area of \Delta ABC}{Area of \Delta DEF}=\frac{BC^2}{EF^2}\]
Construction: Draw \[AL\perp BC\] and \[DM\perp EF\]
Proof:
| STATEMENT |
REASON |
| Compare \[ \Delta ALB \]and\[ \Delta DME\] |
|
| \[\angle ABL=\angle DEM\] |
(\[\because Data\]) |
| \[\angle ALB=\angle DME={90}^{\circ}\] |
(\[\because Construction\]) |
| \[\therefore \Delta ALB \approx \Delta DME\] |
(\[\because Equiangular\]) |
| \[\Rightarrow \frac{AL}{DM}=\frac{AB}{DE}\] |
(\[\because AA criteria\]) |
| but \[\frac{BC}{EF}=\frac{AB}{DE}\] |
(\[\because Data\]) |
| \[\frac{AL}{DM}=\frac{BC}{EF}\] |
(\[\because Transitive property\] |
Area of ABC=\[\frac{1}{2}\times BC\times AL\]
Area of DEF=\[\frac{1}{2}\times EF\times DM\]
\[\therefore \frac{Area of ABC}{Area of DEF}=\frac{\frac{1}{2}\times BC\times AL}{\frac{1}{2}\times EF\times DM}=\frac{BC\times AL}{EF\times DM}\]
Now,\[\frac{ar( \Delta ABC)}{ar( \Delta DEF)}=\frac{BC\times AL}{EF\times DM}\]
\[=\frac{BC}{EF}\times \frac{AL}{DM}\]
\[=\frac{BC}{EF}\times \frac{BC}{EF}\] [\[\because \frac{AL}{DM}=\frac{BC}{EF} \] is proved]
\[\frac{BC^2}{EF^2} \]
\[\therefore \frac{ar( \Delta ABC)}{ar( \Delta ABC)}=\frac{BC^2}{EF^2} \]
From data, \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}\]
\[\therefore \frac{ar( \Delta ABC)}{ar( \Delta DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}\]
