Ceva's Theorem

[logistic_guy]

Prove Ceva’s Theorem: If \(\displaystyle P\) is any point inside \(\displaystyle \Delta ABC\), then \(\displaystyle \frac{AY}{YC} \cdot \frac{CX}{XB} \cdot \frac{BZ}{ZA} = 1\).

Hint: Draw lines parallel to \(\displaystyle \overline{BY}\) through \(\displaystyle A\) and \(\displaystyle C\). Apply theorem \(\displaystyle 6.4\) to \(\displaystyle \Delta ACM\). Show that \(\displaystyle \Delta APN \sim \Delta MPC, \Delta CXM \sim \Delta BXP,\) and \(\displaystyle \Delta BZP \sim \Delta AZN\).

 

[khansaheb]

Prove Ceva’s Theorem: If \(\displaystyle P\) is any point inside \(\displaystyle \Delta ABC\), then \(\displaystyle \frac{AY}{YC} \cdot \frac{CX}{XB} \cdot \frac{BZ}{ZA} = 1\).

Hint: Draw lines parallel to \(\displaystyle \overline{BY}\) through \(\displaystyle A\) and \(\displaystyle C\). Apply theorem \(\displaystyle 6.4\) to \(\displaystyle \Delta ACM\). Show that \(\displaystyle \Delta APN \sim \Delta MPC, \Delta CXM \sim \Delta BXP,\) and \(\displaystyle \Delta BZP \sim \Delta AZN\).

View attachment 39053

show us your effort/s to solve this problem.

 

[logistic_guy]

The diagram already helped us by drawing the three parallel segments \(\displaystyle \text{AN}, \text{YB}\) and \(\displaystyle \text{CM}\).

When we connect the segments \(\displaystyle \text{AM}\) and \(\displaystyle \text{CN}\) through the point \(\displaystyle \text{P}\), we get the following:

\(\displaystyle m\angle \text{APN} = m\angle \text{CPM}\) since they are vertical angles.

And

\(\displaystyle m\angle \text{ANP} = m\angle \text{MCP}\) since they are alternate interior angles.

Then,

\(\displaystyle \Delta \text{APN} \sim \Delta \text{MPC}\)

 

[santo3vong]

Find Local Ladies Ready to Spice Up the Night in Your City

Best Sex Game

 

[logistic_guy]

Since the segment \(\displaystyle \text{YP}\) touches two legs in \(\displaystyle \Delta \text{ACM}\) and parallel to the third leg, by the Triangle Proportionality Theorem (Theorem \(\displaystyle 6.4\)), we have the following:

\(\displaystyle \frac{\text{AY}}{\text{YC}} = \frac{\text{AP}}{\text{PM}}\)