Deriving Menelaus's theorem from Ceva's theorem | plane geometry | intermediate level

Episode 85.

Deriving Menelaus's theorem from Ceva's theorem | plane geometry | intermediate level.

Branch of mathematics: plane geometry.
Difficulty level: intermediate.

Ceva's theorem. Let $ABC$ be a triangle. Let $D$, $E$, $F$ be points on the lines $BC$, $CA$, $AB$ respectively. Then the lines $AD$, $BE$, $CF$ intersect at a single point or all 3 are parallel to each other if and only if $\frac{\overrightarrow{BD}}{\overrightarrow{DC}} \cdot \frac{\overrightarrow{CE}}{\overrightarrow{EA}} \cdot \frac{\overrightarrow{AF}}{\overrightarrow{FB}} = 1$.

Menelaus's theorem. Let $ABC$ be a triangle. Let $D$, $E$, $F$ be points on the lines $BC$, $CA$, $AB$ respectively. Then the points $D$, $E$, $F$ are collinear if and only if $\frac{\overrightarrow{BD}}{\overrightarrow{DC}} \cdot \frac{\overrightarrow{CE}}{\overrightarrow{EA}} \cdot \frac{\overrightarrow{AF}}{\overrightarrow{FB}} = -1$.

This video contains the proof of Menelaus's theorem using Ceva's theorem.

Mathematics. Geometry. Plane geometry.
#Mathematics #Geometry #PlaneGeometry

The same video on YouTube:
https://youtu.be/VqCUzxceodM

The same video on Telegram:
https://t.me/mathematical_bunker/109