Example 25. In $\triangle ABC$, $[BD]$ is a median, point $P \in [BD]$ and $|BP|:|PD|=3:1$; $(AP) \cap [BC]=K$. Find the ratio of the area of $\triangle ABK$ to the area of $\triangle ACK$.
S olution. Let $(A P) \cap(B C)=K$ (Fig. 29). We need to find $S_{\triangle A B K}: S_{\triangle A C K}$. Drop perpendiculars $D M_{1}, B M_{2}, C M_{3}$ to the line $A K$. From the similarity of triangles $B P M_{2}$ and $D P M_{1}$, we write: $\left|B M_{2}\right|:\left|D M_{1}\right|=3: 1$, and from the similarity of triangles $C K M_{3}$ and $D P M_{1}$, we have: $\left|C M_{3}\right|:\left|D M_{1}\right|=2: 1$. Then $\left|B M_{2}\right|:\left|C M_{3}\right|=3: 2$ and, consequently,

$$
\frac{s_{\triangle A B K}}{S_{\triangle A C K}}=\frac{0.5|A K|:\left|B M_{2}\right|}{0.5|A K|:\left|C M_{3}\right|}=\frac{3}{2}
$$

R emark. In the considered example, the discussion is about dependencies in a triangle of arbitrary shape. In problems of type $M$ (and even more so of type $F$), the given and sought properties can only be those that are preserved under similarity transformations.

Therefore, the problem considered in Example 25 is not a problem of type $M$.

At the same time, the property of a figure being a triangle is preserved under affine transformations. Under affine transformations, all other properties of the given triangle in this example are also preserved.

Thus, the considered problem is affine. In the study of affine geometry, such problems were solved by applying affine transformations. In the high school geometry course, affine transformations are not studied. Nevertheless, some problems of affine content are present in the school geometry course. In this case, as in Problem 25, they are solved without the use of affine transformations.

With this, we will conclude the consideration of general methods for solving problems of type $F$ and type $M$ and move on to the last point of the plan for solving a geometric problem - to the investigation.

## § 5. Investigation of the Solution

When solving a geometric problem by an algebraic method, values of the sought quantity may be obtained that do not satisfy the conditions of the problem. For example, the length of a certain segment may turn out to be expressed by a negative number, the angle between two lines may turn out to be greater than $180^{\circ}$, the value of the radius of a circle inscribed in a triangle may turn out to be greater than the side of this triangle, etc.

A value of the sought quantity that does not satisfy the conditions of the problem is called extraneous. Determining which of the found values of the sought quantity satisfy the conditions of the problem and which are extraneous is carried out during the investigation of the solution.

In the process of solving a geometric problem, we consider various figures taking into account their mutual arrangement, perform certain constructions (basic and additional). For what values of parameters (letter data) and for what dependencies between them do the considered figures exist, are the required constructions feasible, what are the variants of the mutual arrangement of the considered figures, and how do all these dependencies affect the existence of a solution and the number of solutions - answers to all these questions are given in the investigation.

Thus, in Example 9 (p. 22), $x$ is the length of a segment, i.e., $x>0$. Solving the equation, we found that $x=\frac{2 a b}{a+b} \cos \frac{\varphi}{2}$. To assert that $|C D|=\frac{2 a b}{a+b} \cos \frac{\varphi}{2}$, it is necessary to ensure the validity of the inequality $\frac{2 a b}{a+b} \cos \frac{\varphi}{2}>0$. Obtaining this inequality and proving its validity constitute the content of the investigation of the solution of this problem.

In the investigation of Example 10 (p. 23), it was required to notice that by the meaning of the problem, the value of $x$ must satisfy the system of inequalities $0<\frac{3}{2}$, and to ensure that the found value of $x$ satisfies this inequality.

Extraneous solutions can also be obtained in the process of solving the composed equation (or system of equations) and even in the process of its composition.

Thus, when solving the system composed of equations (13.1), (13.2), and (13.3) (p. 27), we obtained the equation

$$
\frac{c^{2}}{3} \cos ^{2} t+c^{2} \sin ^{2} 2 t=c^{2}
$$

This equation, together with equations (13.2) and (13.3), formed a new system. It was the solutions of this new system that were sought further. However, this new system of equations is only a consequence of the original system (i.e., it may have extraneous solutions), and therefore its solutions needed to be verified. The investigation allowed us to notice that the value $\cos 2 t=-\frac{1}{3}$ is an extraneous solution.

Sometimes it is convenient to investigate the found value not of the main but of an auxiliary unknown, as was done in Example 13 (p. 26).

Let us return now to Example 15 (p. 29). Solving equation (15.1) $\frac{1}{\sin \frac{x}{2}}=\frac{\sqrt{2}}{\cos x}$, we obtained an extraneous value by the meaning of the problem:

$$
\sin x=\frac{-\sqrt{2}-\sqrt{10}}{4} \text {. }
$$

How did this value come about?

The fact is that we actually solved not one equation (15.1), but a system of equations:

$$
\left\{\begin{array}{l}
|S D|=\frac{a}{2 \sin \frac{x}{2}} \\
|S D|=\frac{a \sqrt{2}}{2 \cos x}
\end{array}\right.
$$

Eliminating $|S D|$ from this system, i.e., already in the process of composing equation (15.1), we expanded the domain of values of $\sin \frac{x}{2}$ and $\cos x$. Indeed, from the equations

$$
|S D|=\frac{a}{2 \sin \frac{x}{2}} \text { and }|S D|=\frac{a \sqrt{2}}{2 \cos x}
$$

it is clear that $\sin \frac{x}{2}>0$ and $\cos x>0$. From equation (15.1), we obtain that

$$
\left\{\begin{array} { l } 
{ \operatorname { s i n } \frac { x } { 2 } > 0 , } \\
{ \operatorname { c o s } x > 0 , }
\end{array} \text { or } \left\{\begin{array}{l}
\sin \frac{x}{2}<0 \\
\cos x<0
\end{array}\right.\right.
$$

This expansion of the domain of values of $\sin \frac{x}{2}$ and $\cos x$, allowed in the process of composing equation (15.1), led to the appearance of an extraneous value of $\sin \frac{x}{2}$.

To perform the investigation, one finds in the problem statement the restrictions that the value of the sought quantity must satisfy. In this case, if in problems containing only numerical data, the restrictions are numerical inequalities (or systems of inequalities), then in problems with letter data (parameters), the restrictions also contain parameters, and thus in such cases, during the investigation, it is necessary to determine the permissible values of the parameters by the meaning of the problem, and sometimes the permissible dependencies between the parameters.

Solving the equation (system of equations) composed according to the conditions of the problem, we often obtain several values of the sought quantity. Thus, in Example 16 (p. 30), two values of $x$ were obtained, i.e., there arises the assumption that in addition to the circle $\omega_{2}$, there is another "unaccounted" circle $\omega_{3}$.

The existence of both these circles can be confirmed by establishing that the found values of $x$ must satisfy the system of inequalities $0<x<R$, and then showing that the found values of $x$ satisfy this system of inequalities.

In Chapters II and III, we will provide several more examples where we will conduct a detailed investigation. Having outlined the solutions, we will focus in detail only on their investigation.