Question

In `DeltaABC`, `/_C` is an acute angle, seg `AD bot` seg `BC`. Prove `AB^(2)=BC^(2)+AC^(2)-2BCxxDC` by completing the following activity.

Let `AB=c`, `AC=b`, `AD=p`, `BC=a`, `DC=x`
`:.BD=a-x`
In `DeltaADB`, by Pythagoras theorem,
`c^(2)=(a-x)^(2)+square`
`c^(2)=a^(2)-2ax+x^(2)+square`...........`(1)`
In `DeltaADC`, by Pythagoras theorem,
`b^(2)=p^(2)+square`
`p^(2)=b^(2)-square` ..........`(2)`
Substituting value of `p^(2)` from `(2)` in `(1)`
`c^(2)=a^(2)-2ax+x^(2)+square`
`c^(2)=a^(2)+b^(2)-square`
`:.AB^(2)=BC^(2)+AC^(2)-2BCxxDC`.

Answer 1

In `DeltaADB`, by Pythagoras theorem,
`c^(2)=(a-x)^(2)+p^(2)`
`c^(2)=a^(2)-2ax+x^(2)+p^(2)`.......`(1)`
In `DeltaADC`, by Pythagoras theorem,
`b^(2)=p^(2)+x^(2)`
`p^(2)=b^(2)-x^(2)`..........`(2)`
Substituting value of `p^(2)` from `(2)` in `(1)`
`c^(2)=a^(2)-2ax+x^(2)+b^(2)-x^(2)`
`c^(2)=a^(2)+b^(2)-2ax`
`:.AB^(2)=BC^(2)+AC^(2)-2BCxxDC`.