In a triangle ABC, let $\angle \mathrm { B } = 45 ^ { \circ }$ and $\angle \mathrm { C } = 75 ^ { \circ }$, and let D be the intersection of the bisector of $\angle \mathrm { A }$ and side BC. (1) From the law of sines we have $$\mathrm { AC } = \frac { \sqrt { \mathbf { A } } } { \sqrt { \mathbf { B } } } \mathrm { BC } , \quad \mathrm { AD } = \sqrt { \mathbf { C } } \mathrm { BD } .$$ In particular, from $\angle \mathrm { ADC } = \mathbf { D E } ^ { \circ }$ we see that $$\mathrm { BD } : \mathrm { BC } = \mathbf { F } : \sqrt { \mathbf { G } }$$ and hence we have $$\mathrm { AB } : \mathrm { AC } = \mathbf { H } : \left( \sqrt { \mathbf { I } } - \frac { \mathbf { J } }{\mathbf{J} } \right) .$$ (2) Let $\mathrm { O } _ { 1 }$ be the center of the circumscribed circle of triangle ABD, and let $\mathrm { O } _ { 2 }$ be the center of the circumscribed circle of triangle ADC. Let us find the ratio of the areas of triangle ABC and triangle $\mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$, $\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$. Since $\angle \mathrm { AO } _ { 1 } \mathrm { D } = \mathbf { K } \mathbf { L } ^ { \circ }$ and $\angle \mathrm { AO } _ { 2 } \mathrm { O } _ { 1 } = \mathbf { M N } ^ { \circ }$, by the same reasoning as (1), we have $$\mathrm { AC } = \sqrt { \mathbf { O } } \mathrm { AO } _ { 1 } , \quad \mathrm { AO } _ { 2 } = ( \sqrt { \mathbf { P } } - \mathbf { Q } ) \mathrm { AO } _ { 1 } .$$ Hence we obtain $$\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 } = \mathbf { R } : ( \mathbf { S } - \sqrt { \mathbf { T } } ) .$$
In a triangle ABC, let $\angle \mathrm { B } = 45 ^ { \circ }$ and $\angle \mathrm { C } = 75 ^ { \circ }$, and let D be the intersection of the bisector of $\angle \mathrm { A }$ and side BC.\\
(1) From the law of sines we have
$$\mathrm { AC } = \frac { \sqrt { \mathbf { A } } } { \sqrt { \mathbf { B } } } \mathrm { BC } , \quad \mathrm { AD } = \sqrt { \mathbf { C } } \mathrm { BD } .$$
In particular, from $\angle \mathrm { ADC } = \mathbf { D E } ^ { \circ }$ we see that
$$\mathrm { BD } : \mathrm { BC } = \mathbf { F } : \sqrt { \mathbf { G } }$$
and hence we have
$$\mathrm { AB } : \mathrm { AC } = \mathbf { H } : \left( \sqrt { \mathbf { I } } - \frac { \mathbf { J } }{\mathbf{J} } \right) .$$
(2) Let $\mathrm { O } _ { 1 }$ be the center of the circumscribed circle of triangle ABD, and let $\mathrm { O } _ { 2 }$ be the center of the circumscribed circle of triangle ADC. Let us find the ratio of the areas of triangle ABC and triangle $\mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$, $\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 }$.
Since $\angle \mathrm { AO } _ { 1 } \mathrm { D } = \mathbf { K } \mathbf { L } ^ { \circ }$ and $\angle \mathrm { AO } _ { 2 } \mathrm { O } _ { 1 } = \mathbf { M N } ^ { \circ }$, by the same reasoning as (1), we have
$$\mathrm { AC } = \sqrt { \mathbf { O } } \mathrm { AO } _ { 1 } , \quad \mathrm { AO } _ { 2 } = ( \sqrt { \mathbf { P } } - \mathbf { Q } ) \mathrm { AO } _ { 1 } .$$
Hence we obtain
$$\triangle \mathrm { ABC } : \triangle \mathrm { AO } _ { 1 } \mathrm { O } _ { 2 } = \mathbf { R } : ( \mathbf { S } - \sqrt { \mathbf { T } } ) .$$