Ceva's Theorem

Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.

Prove the following


First we need to construct a parallel line to BC through A. Extend BE to point Z and CF to Y so that Y and Z lie on this parallel line.



In order to explore the ratio, we should find some similar triangles (ratio ==> similar triangle ==> parallel lines; this is the thought process).
Angle APY = Angle CPD because vertical angles are congruent.

Angle AYP = Angle PCD because alternate interior angles are congruent.

Angle YAP = Angle CDP """

Triangle YAP ~ Triangle CDP by AAA.

Then


Similarly, triangle ZAP ~ triangle BDP by AAA.

Then



By substitution,

By nature of ratios, then



By the same construction and properties therein, triangle ZAE ~ triangle BCE.

And



And triangle YAF ~ triangle CBF,

so


Just to make things easier to see, here are all the important ratios together.

, ,

Since each is equal to another, let's take one side from each of the three equalities and multiply them.

Take the other side of each of the three equalities and multiply , then the ratios are equal.

=

As you can see, the right side of this equation reduces to 1.

= 1

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