Suppose that in the figure to the right $$\mathrm { AB } = 4 , \quad \mathrm { AC } = 5 , \quad \cos \angle \mathrm { BAC } = \frac { 1 } { 8 }$$ and $$\angle \mathrm { BAD } = \angle \mathrm { ACB } , \quad \angle \mathrm { CAE } = \angle \mathrm { ABC } .$$ (1) When we denote the area of $\triangle \mathrm { ABC }$ by $S$, we have $$S = \frac { \square \mathbf { A B } \sqrt { \square \mathbf { C } } } { \square } .$$ Also $\mathrm { BC } = \mathbf { E }$. (2) Furthermore, when we denote the areas of $\triangle \mathrm { ABD }$ and $\triangle \mathrm { ACE }$ by $S _ { 1 }$ and $S _ { 2 }$, respectively, we have $$S : S _ { 1 } : S _ { 2 } = 1 : \frac { \mathbf { F } } { \mathbf{G} } : \frac { \mathbf { H I } } { \mathbf { J } } .$$ (3) When we denote the area of $\triangle \mathrm { ADE }$ by $T$, we have $$T = \frac { \mathbf { L M } \sqrt { \mathbf { N } } } { \mathbf { O P } } .$$ Also $\mathrm { DE } = \dfrac { \mathbf { Q } } { \mathbf{R} }$.
Suppose that in the figure to the right
$$\mathrm { AB } = 4 , \quad \mathrm { AC } = 5 , \quad \cos \angle \mathrm { BAC } = \frac { 1 } { 8 }$$
and
$$\angle \mathrm { BAD } = \angle \mathrm { ACB } , \quad \angle \mathrm { CAE } = \angle \mathrm { ABC } .$$
(1) When we denote the area of $\triangle \mathrm { ABC }$ by $S$, we have
$$S = \frac { \square \mathbf { A B } \sqrt { \square \mathbf { C } } } { \square } .$$
Also $\mathrm { BC } = \mathbf { E }$.
(2) Furthermore, when we denote the areas of $\triangle \mathrm { ABD }$ and $\triangle \mathrm { ACE }$ by $S _ { 1 }$ and $S _ { 2 }$, respectively, we have
$$S : S _ { 1 } : S _ { 2 } = 1 : \frac { \mathbf { F } } { \mathbf{G} } : \frac { \mathbf { H I } } { \mathbf { J } } .$$
(3) When we denote the area of $\triangle \mathrm { ADE }$ by $T$, we have
$$T = \frac { \mathbf { L M } \sqrt { \mathbf { N } } } { \mathbf { O P } } .$$
Also $\mathrm { DE } = \dfrac { \mathbf { Q } } { \mathbf{R} }$.