CPM Homework Help : CCGS Problem 5-23

Examina los triángulos de la derecha. ¿Los triángulos son semejantes? Si así fuera, muestra cómo lo sabes con un diagrama de flujo. Caso contrario, explica cómo sabes que no pueden ser semejantes.

Two decreasing parallel lines with a transversal segment between them, A, E intersecting the right parallel line at, C, and the left parallel line at D. A point B above and on the left parallel line is part of triangle, A, B, C where B, C is 4 and A, C is 6. A point F below and on the right parallel line is part of triangle, D, E, F where D, E is 7.5 and D, F is 5.

Paso 1:

Asegúrate de agregarle al diagrama de flujo los enunciados lógicos y las flechas.

Paso 2:

Paso 3:

Paso 4:

$\begin{array}{c} \qquad \qquad \quad \enclose{circle}{\; \\ \quad \angle BCA \cong \angle DCX \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \angle DCX \cong \angle EDF \quad \\} \end{array}$

$\begin{array}{c} \enclose{circle}{\; \\ \quad \angle BCA \cong \angle DCX \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \angle DCX \cong \angle EDF \quad \\} \\ \enclose{circle}{\; \\ \quad \frac{BC}{ED} = \frac{4}{6} = \frac{2}{3} \quad \\} \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \enclose{circle}{\; \\ \quad \frac{AC}{DF} = \frac{5}{7.5} = \frac{2}{3} \quad \\} \\ \end{array}$

$\begin{array}{c} \enclose{circle}{\; \\ \quad \angle BCA \cong \angle DCX \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \angle DCX \cong \angle EDF \quad \\} \\ \enclose{circle}{\; \\ \quad \frac{BC}{ED} = \frac{4}{6} = \frac{2}{3} \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \angle BCA \cong \angle EDF \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \frac{AC}{DF} = \frac{5}{7.5} = \frac{2}{3} \quad \\} \\ \end{array}$

$\begin{array}{c} \enclose{circle}{\; \\ \quad \angle BCA \cong \angle DCX \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \angle DCX \cong \angle EDF \quad \\} \\ \enclose{circle}{\; \\ \quad \frac{BC}{ED} = \frac{4}{6} = \frac{2}{3} \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \angle BCA \cong \angle EDF \quad \\} \qquad \qquad \enclose{circle}{\; \\ \quad \frac{AC}{DF} = \frac{5}{7.5} = \frac{2}{3} \quad \\} \\ \enclose{circle}{\; \\ \quad \Delta BCA \sim \Delta EDF \quad \\} \end{array}$