User:Extra999/sandbox

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

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

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.