Relating areas and angles using Ceva’s theorem and Menalaus’ theorem


Ceva’s theorem and Menalus’ theorem are widely applied in geometry problems in various ways. As long as the shape that Ceva’s theorem and Menalaus’ theorem is associated with appears in a geometric diagram, it is often useful to write down the expression of these theorem applied to the question and see if it of any use. However, these theorems are not only useful in relating the ratios of lengths. Consider the following problem from the recent APMO.

(APMO 2012 P1) Let P be a point in the interior of a triangle ABC, and let D, E, F be the points of intersection of the line AP and the side BC of the triangle, of the line BP and the side CA, and of the line CP and the side AB respectively. Prove that the area of the triangle ABC must be 6 if the area of each of the triangles PFA, PDB and PEC is 1.

This is a very popular diagram that often comes out in MO problems. There are several ratios involved in this diagram that makes it so accessible to problem setters. In particular, this triangle can form 6 Menalaus theorem relationships. To solve the problem above, we shall use the following three equations:

\dfrac{BD}{CD}\cdot\dfrac{CP}{PF}\cdot\dfrac{AF}{AB}=1

\dfrac{AF}{Fb}\cdot\dfrac{BP}{PE}\cdot\dfrac{EC}{AC}=1

\dfrac{CE}{EA}\cdot\dfrac{AP}{PD}\cdot\dfrac{BD}{BC}=1

However, it is not very useful to find the ratios of lengths since we are not given any details about the segments in the triangles. However, we can use Menalaus’ theorem to relate the area of triangles. We let x be the area of triangle PCD, y be the area of triangle PEA and z be the area of triangle PFB. Replacing the ratio of lengths using the ratios of areas, we have the following system of simultaneous equations:

\dfrac{1}{x}\cdot\dfrac{1+y}{1}\cdot\dfrac{1}{1+z} =1

\dfrac{1}{z}\cdot\dfrac{1+x}{1}\cdot\dfrac{1}{1+y} =1

\dfrac{1}{y}\cdot\dfrac{1+z}{1}\cdot\dfrac{1}{1+x} =1

There are many ways to solve this simultaneous equation. It is possible to solve the system of equations using inequalities too. By rearranging and adding the equations in the system, we obtain xy+yz+xz=3. Using Ceva’s theorem to relate areas, we also have xyz=1. Using AM-GM inequality,  we know that 1=\sqrt[3]{x^2y^2z^2} \le \dfrac{xy+yz+xz}{3} =1. Since the equality case is only established when all terms are equal, we have x=y=z=1 and hence the area of the large triangle is 6.


Ceva’s theorem can also be used to relate angles too. From the above diagram, we also have:

\dfrac{\sin{\angle BAD}}{\sin{\angle CAD}}\cdot\dfrac{\sin{\angle ACF}}{\sin{\angle BCF}} \cdot \dfrac{\sin{\angle CBE}}{\sin{\angle ABE}}=1

Using this form of Ceva’s theorem, we can easily prove the three angle bisectors of a triangle are concurrent i.e. intersect at a common point. The following problem form IMO Shortlist 2001 yields a very beautiful application of Ceva’s theorem in its angular form:

Let A_1 be the center of the square inscribed in acute triangle ABC with two vertices of the square on side BC. Thus one of the two remaining vertices of the square lies on side AB and the other on segment AC. Points B_1 and C_1 are defined in a similar way for inscribed squares with two vertices on sides AC and AB, respectively. Prove that lines AA_1, BB_1, CC_1 are concurrent.

Sine rule helps in solving the above problem. Do give it a try.

Cheers,

ksj

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