Theorem: In a right angled triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.
Data: In \[ \Delta ABC, \angle ABC= {90}^{\circ}\]
To prove: \[AB^2+BC^2= CA^2\]
Construction: Draw \[BD \perp AC.\]
Proof:
| STATEMENT |
REASON |
| \[Compare \Delta ABC and \Delta ADB\] |
|
| \[\angle ABC= \angle ADB= {90}^{\circ}\] |
(\[\because \]Data and construction) |
| \[\angle BAD \] is common. |
|
| \[\therefore \Delta ABC\approx \Delta ADB\] |
(\[\because \] Equiangular triangles) |
| \[\Rightarrow \frac{AB}{AD}=\frac{AC}{AB}\] |
(\[\because \] AA similarity criteria) |
| \[\therefore AB^2=AC \times AD\].....(1) |
|
| \[Compare \Delta ABCand \Delta BDC.\] |
|
| \[\angle ABC=\angle BDC= {90}^{\circ}\] |
(\[\because \]Data and construction) |
| \[\angle ACB\] is common. |
|
| \[\therefore \Delta ABC \approx \Delta BDC\] |
(\[\because \] Equiangular triangles) |
| \[\Rightarrow \frac{BC}{DC}=\frac{AC}{BC}\] |
(\[\because \] AA similarity criteria) |
| \[BC^2=AC\times DC\]......2 |
|
By adding (1) and (2) we get,
\[AB^2+BC^2=(AC\times AD)+(AC\times DC)\]
\[AB^2+BC^2=AC(AD+DC)\]
\[AB^2+BC^2=AC\times AC=AC^2\] \[(\because AD+DC=AC)\]
\[\therefore AB^2+BC^2=AC^2\]
