Constructing a line that passes through $P$ I have recently read a book by Heisuke Hironaka. However, the book is not available on English. The book was basically a biography on his life. Heisuke Hironaka says that his high school teacher had given him a problem only he had managed to solve. 
"Construct a line that passes through $P$ that meets lines $AB$,$AC$, on points $K$, $M$ so that $BK=CM$."
So far, I have found out that if the perpendicular bisecor of $BC$ and the external angle bisector of $A$ met on the point $L$, for the condition $BK=CM$ to be satisfied, $LK=LM$. Also, if the line that passed through $P$ met the external angle bisector of $A$ on the point $T$, the angle $LTP$ is equal to the angle $LKA$. However, I was able to progress no further from here. 
Any help would be appreciated. 
 A: I will present a construction in two steps. There are a number of exceptional cases to be considered, but they should be ignored on first reading.
We will assume $A, B, C$ are not aligned. (Otherwise, the only solution is to let $K = M$ be the midpoint of $BC$.) 
First note that there are two rotations $\rho$ which carry line $AB$ to line $AC$ and which carry $B$ to $C$. For each point $K$ of line $AB$, the corresponding point $M = \rho(K)$ satisfies $CM = BK$. Therefore the only problem is how to select $K$ so that $P$, $K$ and $M$ are aligned. In fact, all solutions are obtained this way for one of the two rotations $\rho$.
The first step in the construction is to find the centre $F$ of one of the rotations. This can be done by letting $D = \rho(A)$ be either of the two points on $AC$ satisfying $CD = BA$, and taking $F$ to be the intersection of the perpendicular bisectors of segments $BC$ and $AD$, since $FB = FC, FA=FD$. (If $D = A$, let $F = A$.) Thus $\rho$ is the rotation with centre $F$ carrying $B$ to $C$.
Now for any point $K$ on $AB$ and $M = \rho(K)$ on $AC$, let $I$ be the midpoint of $KM$. As $K$ varies, it is not hard to see that the locus of $I$ is a line $a$. The line $a$ can easily be constructed by selecting two arbitrary points $K_1$ and $K_2$ on $AB$ and drawing the corresponding line $I_1 I_2$. (For example, if $K_1 = B, M_1 = C, K_2 = A, M_2 = D$, then the points $I_1$ and $I_2$ have already been constructed!) Conversely, if the midpoint $I$ of $KM$ is given, then the points $K$ and $M$ can be obtained as the intersections with $AB$ and $AC$ of the line through $I$ perpendicular to $FI$, since $I$ is the midpoint of a chord $KM$ in a circle centred at $F$. (An exceptional case occurs when $F = A = I$; in that case, $K = M = A$. There will be further exceptional cases where the chord coincides with $AB$ or $AC$; but at least one of the points $K$, $M$ can always be obtained as an intersection, and then the other as the reflection of the first with respect to $I$.)
If we let $I$ be any point on line $a$, then the corresponding points $K$ and $M$ answer the question if and only if $P$ belongs to line $KIM$. This is equivalent to $\angle FIP$ being a right angle, hence to $I$ belonging to the circle with diameter $FP$. 
Therefore the last step in the construction is to take any point of intersection $I$ of line $a$ with the circle with diameter $FP$. Then $PI$ is the required line. (If, exceptionally, $P = I$, then any line through $P$ will do.)
In summary (general case)


*

*Choose $D$ on $AC$ so that $CD = BA$.

*Construct the perpendicular bisectors $l_1$ and $l_2$ of segments $BC$ and $AD$. Let $F$ be their intersection.

*Let $I_1$, $I_2$ be the midpoints of segments $BC$ and $AD$.

*Let $I$ be a point of intersection of line $I_1I_2$ with the circle with diameter $FP$.

*$PI$ is the required line. 

A: This is NOT a solution. I am just sharing my finding.

1) The problem would be extremely difficult when P is just a random given point on the plane.
1.1) It is pointless to assume that ABC is isosceles with AB = AC. This is because h, the external angle bisector of angle A, is then the required line.
1.2) Then, we can assume that P is not on h but PKM will cut h at some point P’. That is, P’KM is our target line instead (which of course is a bit easier).
Following your logic, we can 
(i) construct line k, the perpendicular bisector of BC; and let it cut BA extended at X;
(ii) if h cuts k at L, let circle $\pi$ [center $= L$ & radius $= LX$] cut $AB$ and $AC$ at $K_1$ and $M_1$ respectively.
$TK_1M_1$ is the required line for a special case solution only. $P’ (= T$ this time) lying on h and the line cuts the circle $\pi $ at $K_1$ and $M_1$). This can be proved by considering congruent triangles $K_1BL$ and $M_1CL$ as mentioned.
Note that $TXLK_1$ forms a cyclic quadrilateral for this special case only. Other points like P’ does not have this property.
The general case is when $P’$ is not on the circle $\pi $. Again by considering congruent triangles, we know that the required $K$ and $M$ will both lie on one of the dotted concentric circles (center $= L$) whose radius is not known because of insufficient information on $P$.
