User:Extra999/sandbox

User:Extra999/sandbox at Pr$\infty$fWiki

To do: Gabba (improve fortress para), Sudha dairy provide links and fix history

$T={\frac {1}{4}}{\sqrt {4a^{2}b^{2}-(a^{2}+b^{2}-c^{2})^{2}}}$

Using the properties of the tangents to circles, we know that AF = AE, EC = ED, FB = DB, and ang(OFA) = ang(OEA) = ang(ODC) = 90°. So (BD + DC) + (BF + AF) + (AE + EC) = 2(AF + EC + FB) = AB + BC + AC = 2s. Now that BH = AF, we have CH = s. Using the area formula, rs, the area of the triangle can be written as (CH)(OD).

Using the properties of the tangents to circles, we know that $AF=AE$, $EC=ED$, $FB=DB$, and $\angle {OFA}=\angle {OEA}=\angle {ODC}=90\,^{\circ }\mathrm {}$. So $(BD+DC)+(BF+AF)+(AE+EC)=2(AF+EC+FB)=AB+BC+AC=2s$. Now that $BH=AF$, we have $CH=s$. Using the area formula, $rs$, the area of the triangle can be written as $(CH)(OD)$.

Since $\angle {CBL}=\angle {COL}=90\,^{\circ }\mathrm {}$ and share a common hypotenuse $CL$, and thus follows that they are inscribed in a common circle with $CL$ as the diameter. Therefore, $LBOC$ is a cyclic quadrilateral and hence $\angle {BOC}+\angle {BLC}=180\,^{\circ }\mathrm {}$.

We know that the angle around point O is 360°, that is 2(ang(AOE) + ang(COD) + ang(BOF)) = 360°. Therefore ang(AOE) + ang(BOC) = 180°. Using our previous result, we can say that ang(BLC) = ang(AOE).

Therefore, triangles AOE and BLC are similar by AA criterion. As, HB = AF, so

BC/HB= BC/AF

Using the definition of similarity, we have BC/AF = BL/OE. Now, ang(LKB) = ang(DKO) as they are vertically opposite, and ang(LBH) = ang(ODK) = 90° by construction. Therefore, triangles LKB and DKO are also similar by AA criterion. Thus, BL/OD = BK/KD. As OE = OD = r, we have the important result BC/HB = BK/KD before some algebra follows.