Three Square Problem

Type on YouTube "The Three Square Geometry Problem - Numberphile"

$\frac{AF}{FG}=\sqrt{2}$

$\frac{AH}{AG}=\frac{\sqrt{10}}{\sqrt{5}}=\sqrt{2}$

$\frac{FH}{AF}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$\triangle{AFH}$ is similar to $\triangle{AFH}$ then $\widehat{FAG}=\gamma$

$\beta=\widehat{GAC}$

$\alpha=\widehat{FAE}$

$\alpha + \beta +\gamma =\widehat{FAE}+\widehat{GAC}+\widehat{FAG}=90$

$\angle$ AGH = 45 (DIAGONALS BISECT THE ANGLES = 90/2).....

tan $\angle$ AFH= 1/2.....

tan $\angle$ AEH=1/3......

tan[ $\angle$ AFH+ $\angle$ AEH]= tan $\angle$ AFH+tan $\angle$ AEH/1-tan $\angle$ AFH tan $\angle$ AEH} = $\frac{1/2+1/3}{1-1/2*1/3}$ = $\frac{5/6}{5/6}$ = 1........

So $\angle$ AFH+ $\angle$ AEH=45........

SO $\angle$ AGH+ $\angle$ AFH+ $\angle$ AEH = 45 + 45 = $\boxed{90}$

Rearrange the triangles as shown. If we let the original square have a side length of $x$ , we can get the lengths shown by given measurements, the Pythagorean Theorem, or the Distance Formula. Note that the quadrilateral satisfies the converse of Ptolemy's Theorem. Therefore, this quadrilateral is a cyclic quadrilateral, so the opposite angles add to $180^\circ$ . This gives $\alpha+\beta+\gamma+90^\circ=180^\circ$ $\alpha+\beta+\gamma=\boxed{90^\circ}.$

Alpha=45 ......angle made by diagonal Tan(beta)=1/2 Tan (gamma)=1/3 Now use property, tan (A+B)=(tanA+tanB) /1-tanAtanB Get the answer