Theorem:
If two triangles are equiangular, then their corresponding sides are proportional.
Data: In \[ \Delta ABC \] and \[ \Delta DEF\]
1) \[ B\hat{A} C= E\hat{D} F\]
2) \[A\hat{B} C=D\hat{E} F\]
3) \[A\hat{C} B=D\hat{F} E\]
To Prove: \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\]
Construction: Mark points G and H on AB and AC such that
1) AG = DE
2) AH = DF.
Join G and H.
Proof:
| STATEMENT |
REASON |
| Compare \[ \Delta AGH\] and \[ \Delta DEF\] |
|
| AG = DE |
\[(\because Construction)\] |
| \[\angle GAH=\angle EDF\] |
\[(\because Data)\] |
| AH = DF |
\[(\because Construction)\] |
| \[\therefore \Delta AGH\cong \Delta DEF\] |
\[(\because SAS)\] |
| \[\therefore \angle AGH=\angle DEF\] |
\[(\because CPCT)\] |
| But \[\angle ABC=\angle DEF\] |
\[(\because Data)\] |
| \[\Rightarrow \angle AGH=\angle ABC\] |
\[(\because Axiom-1)\] |
| \[\therefore GH\left | \right | BC\] |
(\[\because \] If corrsponding angles are equal then lines are parallel.) |
| \[\therefore In \Delta ABC\] \[\frac{AB}{AG}=\frac{BC}{GH}=\frac{CA}{HA}\] |
\[\because \] Third corollary to Thales Theorem |
| Hence \[\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}\] |
\[(\because \Delta AGH= \Delta DEF)\] |
