Question

Let `bar(AB)` be cut into equal and unequal segments at `C and D` Show that the rectangle contained by `bar(AD)xxDB` together with the square on `CD` is equal to the square on `CB`?

Answer 1

In the fig C is mid point of AB . So `AC=BC`

Now rectangle contained by `bar(AD) and bar(DB)` together with the square on`bar(CD)`

`=bar(AD)xxbar(DB)+bar(CD)^2`

`=(bar(AC)+bar(CD))xx(bar(BC)-bar(CD))+bar(CD)^2`

`=(bar(BC)+bar(CD))xx(bar(BC)-bar(CD))+bar(CD)^2`

`=bar(BC)^2-cancel(bar(CD)^2)+cancel(bar(CD)^2)`

`=bar(BC)^2->"Square on CB"` Proved

Answer 2