# 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?

Oct 23, 2016

In the fig C is mid point of AB . So $A C = B C$

Now rectangle contained by $\overline{A D} \mathmr{and} \overline{D B}$ together with the square on$\overline{C D}$

$= \overline{A D} \times \overline{D B} + {\overline{C D}}^{2}$

$= \left(\overline{A C} + \overline{C D}\right) \times \left(\overline{B C} - \overline{C D}\right) + {\overline{C D}}^{2}$

$= \left(\overline{B C} + \overline{C D}\right) \times \left(\overline{B C} - \overline{C D}\right) + {\overline{C D}}^{2}$

$= {\overline{B C}}^{2} - \cancel{{\overline{C D}}^{2}} + \cancel{{\overline{C D}}^{2}}$

$= {\overline{B C}}^{2} \to \text{Square on CB}$ Proved

Oct 29, 2016