mathematics chaleng ( 2 ) - How does the length BFBF relate to the original rectangle?

mathematics chaleng ( 2 )  - How does the length BFBF relate to the original rectangle?

Given rectangle ABCDABCD, ABAB is extended to EE such that BE=BCBE=BC. AEAE forms the diameter of a circle and CBCB is extended to FF which lies on the circle.

How does the length BF$BF$ relate to the original rectangle?












solution :

Consider the following diagram.

As triangle AEF$AEF$ is in a semi-circle, ∠AFE=90⇒∠FAB+∠FEB=90$\mathrm{\angle }AFE=90\Rightarrow \mathrm{\angle }FAB+\mathrm{\angle }FEB=90$

But ∠FAB+∠AFB=90⇒∠AFB=∠FEB$\mathrm{\angle }FAB+\mathrm{\angle }AFB=90\Rightarrow \mathrm{\angle }AFB=\mathrm{\angle }FEB$. In the same way, ∠BFE=∠FAB$\mathrm{\angle }BFE=\mathrm{\angle }FAB$.

Thus the right angled triangles FAB$FAB$ and FEB$FEB$ are similar.

∴ABFB=FBBE⇒AB⋅BE=(FB)2$\therefore {\displaystyle \frac{AB}{FB}}={\displaystyle \frac{FB}{BE}}\Rightarrow AB\cdot BE=(FB{)}^2$

But BE=BC$BE=BC$, so AB⋅BC=(FB)2$AB\cdot BC=(FB{)}^2$

Hence the square on FB$FB$ is equal to the area of the rectangle ABCD$ABCD$ and it can be seen that this construction process has "squared" the rectangle.

Show how this method can be adapted to construct the square root of a given length AB=x$AB=x$.