Problem
As shown in the figure, there are two circles, a larger and a smaller, inside a right triangle, which touch each side.
The two circles are in contact across the oblique line $AD$ drawn from the point $A$ to the side $BC$.
When you know $BD$, $∠ABD$, and $∠ADC$, find the diameters of the larger and smaller circles.
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Solution
Let $BD=a, \ ∠ABD=α$, and $∠ADC=β$.
Let the diameters of the greater and the smaller circles be $d_1$ and $d_2$, respectively.
Let the foot of the perpendicular line drawn from the point $D$ to the side $AB$ be $H$.
For the right-angled triangle $∆BDH$,
$$\frac{DH}{BD}=\frac{DH}{a}=sinα,$$
$$∴ \ DH=a×sinα.\qquad [1]$$
Since $∠DAH=β-α$, for the right-angled triangle $∆ADH$,
$$\frac{DH}{AD}=sin(β-α),$$
$$∴ \ AD=\frac{DH}{sin(β-α)}. \qquad [2]$$
Substituting $[1]$ into $[2]$,
$$AD=\frac{a×sinα}{sin(β-α)}. \qquad [3]$$
For the right-angled triangle $∆ADC$,
$$\frac{DC}{AD}=cosβ,$$
$$DC=AD×cosβ,$$
$$∴ \ DC=\frac{a×sinα×cosβ}{sin(β-α)}. \qquad [4]$$
$$\frac{CA}{AD}=sinβ,$$
$$CA=AD×sinβ,$$
$$∴ \ CA=\frac{a×sinα×sinβ}{sin(β-α)}.$$
Since $d_1=DC+CA-AD$,
$$d_1=\frac{a×sinα×cosβ+a×sinα×sinβ-a×sinα}{sin(β-α)}$$
$$=\frac{a×sinα×(cosβ+sinβ-1)}{sin(β-α)}.$$
Next, regarding the right triangle $∆AGO’$, since $AG=AE$,
$$AG=AC-\frac{d_1}{2}$$
$$=\frac{a×sinα×sinβ}{sin(β-α)} -\frac{a×sinα×(cosβ+sinβ-1)}{2 sin(β-α)}$$
$$=\frac{2a×sinα×sinβ-a×sinα×(cosβ+sinβ-1)}{2 sin(β-α)}$$
$$=\frac{a×sinα×(sinβ-cosβ+1)}{2 sin(β-α)}.$$
Since $∠GAO’=(β-α)/2$,
$$\frac{GO’}{AG}=\frac{2 sin(β-α)}{a×sinα×(sinβ-cosβ+1)}×\frac{d_2}{2}$$
$$=\frac{sin(β-α)×d_2}{a×sinα×(sinβ-cosβ+1)}$$
$$=tan\frac{β-α}{2}$$
$$∴ \ d_2=\frac{a×sinα×(sinβ-cosβ+1)}{sin(β-α)}×tan\frac{β-α}{2}.$$
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Reference
Yoshikazu Yamakawa, ed. (1997) Okayama ken no Sangaku (Sangaku in Okayama Prefecture), pp.44-45; pp.327-328.