Ceva's Theorem

logistic_guy

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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 \(\displaystyle \Delta BZP \sim \Delta AZN\).

challenge.png
 
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.
 
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,

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