In $\Delta GBS$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{GS}$. Prove that $\frac{GB^{2}}{BS^{2}} = \frac{GM}{SM}$.
(A) In $\Delta GBS$,$\angle B = 90^{\circ}$ and $BM \perp GS$.
By the property of similar triangles in a right-angled triangle,when an altitude is drawn to the hypotenuse,the triangles formed on both sides of the altitude are similar to the original triangle and to each other.
Specifically,$\Delta GMB \sim \Delta GBS$ and $\Delta B M S \sim \Delta GBS$.
From $\Delta GMB \sim \Delta GBS$,we have $\frac{GB}{GS} = \frac{GM}{GB}$,which implies $GB^{2} = GM \cdot GS$ (Equation $1$).
From $\Delta BMS \sim \Delta GBS$,we have $\frac{BS}{GS} = \frac{SM}{BS}$,which implies $BS^{2} = SM \cdot GS$ (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{GB^{2}}{BS^{2}} = \frac{GM \cdot GS}{SM \cdot GS} = \frac{GM}{SM}$.
Hence,the result is proved.
By the property of similar triangles in a right-angled triangle,when an altitude is drawn to the hypotenuse,the triangles formed on both sides of the altitude are similar to the original triangle and to each other.
Specifically,$\Delta GMB \sim \Delta GBS$ and $\Delta B M S \sim \Delta GBS$.
From $\Delta GMB \sim \Delta GBS$,we have $\frac{GB}{GS} = \frac{GM}{GB}$,which implies $GB^{2} = GM \cdot GS$ (Equation $1$).
From $\Delta BMS \sim \Delta GBS$,we have $\frac{BS}{GS} = \frac{SM}{BS}$,which implies $BS^{2} = SM \cdot GS$ (Equation $2$).
Dividing Equation $1$ by Equation $2$:
$\frac{GB^{2}}{BS^{2}} = \frac{GM \cdot GS}{SM \cdot GS} = \frac{GM}{SM}$.
Hence,the result is proved.
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