Regarding Hilbert's axioms, I've never understood why I.4 is needed, given I.3. From Wikipedia:

I.1: Two distinct points A and B always completely determine a straight line a. We write AB = a or BA = a. Instead of Ã¢??determine,Ã¢?? we may also employ other forms of expression; for example, we may say Ã¢??A lies upon aÃ¢??, Ã¢??A is a point of aÃ¢??, Ã¢??a goes through A and through BÃ¢??, Ã¢??a joins A and/or with BÃ¢??, etc. If A lies upon a and at the same time upon another straight line b, we make use also of the expression: Ã¢??The straight lines a and b have the point A in common,Ã¢?? etc.

I.2: Any two distinct points of a straight line completely determine that line; that is, if AB = a and AC = a, where B Ã¢?Â C, then also BC = a.

I.3: Three points A, B, C not situated in the same straight line always completely determine a plane Ã?Â±. We write ABC = Ã?Â±. We employ also the expressions: Ã¢??A, B, C, lie in Ã?Â±Ã¢??; Ã¢??A, B, C are points of Ã?Â±Ã¢??, etc.

I.4: Any three points A, B, C of a plane Ã?Â±, which do not lie in the same straight line, completely determine that plane.

I.5: If two points A, B of a straight line a lie in a plane Ã?Â±, then every point of a lies in Ã?Â±. In this case we say: Ã¢??The straight line a lies in the plane Ã?Â±,Ã¢?? etc.

I.6: If two planes Ã?Â±, Ã?Â² have a point A in common, then they have at least a second point B in common.

I.7: Upon every straight line there exist at least two points, in every plane at least three points not lying in the same straight line, and in space there exist at least four points not lying in a plane.

II. Order

II.1: If a point B is between points A and C, B is also between C and A, and there exists a line containing the points A,B,C.

II.2: If A and C are two points of a straight line, then there exists at least one point B lying between A and C and at least one point D so situated that C lies between A and D.

II.3: Of any three points situated on a straight line, there is always one and only one which lies between the other two.

II.5: Pasch's Axiom: Let A, B, C be three points not lying in the same straight line and let a be a straight line lying in the plane ABC and not passing through any of the points A, B, C. Then, if the straight line a passes through a point of the segment AB, it will also pass through either a point of the segment BC or a point of the segment AC.

III. Parallels

III.1: In a plane Ã?Â± there can be drawn through any point A, lying outside of a straight line a, one and only one straight line which does not intersect the line a. This straight line is called the parallel to a through the given point A.

IV. Congruence

IV.1: If A, B are two points on a straight line a, and if AÃ?Â¡ is a point upon the same or another straight line aÃ?Â¡ , then, upon a given side of AÃ?Â¡ on the straight line aÃ?Â¡ , we can always find one and only one point BÃ?Â¡ so that the segment AB (or BA) is congruent to the segment AÃ?Â¡ BÃ?Â¡ . We indicate this relation by writing AB Ã¢?? AÃ?Â¡ BÃ?Â¡. Every segment is congruent to itself; that is, we always have AB Ã¢?? AB.

We can state the above axiom briefly by saying that every segment can be laid off upon a given side of a given point of a given straight line in one and only one way.

IV.2: If a segment AB is congruent to the segment AÃ?Â¡BÃ?Â¡ and also to the segment AÃ?Â¡Ã?Â¡BÃ?Â¡Ã?Â¡, then the segment AÃ?Â¡BÃ?Â¡ is congruent to the segment AÃ?Â¡Ã?Â¡BÃ?Â¡Ã?Â¡; that is, if AB Ã¢?? AÃ?Â¡B and AB Ã¢?? AÃ?Â¡Ã?Â¡BÃ?Â¡Ã?Â¡, then AÃ?Â¡BÃ?Â¡ Ã¢?? AÃ?Â¡Ã?Â¡BÃ?Â¡Ã?Â¡

IV.3: Let AB and BC be two segments of a straight line a which have no points in common aside from the point B, and, furthermore, let AÃ?Â¡BÃ?Â¡ and BÃ?Â¡CÃ?Â¡ be two segments of the same or of another straight line aÃ?Â¡ having, likewise, no point other than BÃ?Â¡ in common. Then, if AB Ã¢?? AÃ?Â¡BÃ?Â¡ and BC Ã¢?? BÃ?Â¡CÃ?Â¡, we have AC Ã¢?? AÃ?Â¡CÃ?Â¡.

IV.4: Let an angle (h, k) be given in the plane Ã?Â± and let a straight line aÃ?Â¡ be given in a plane Ã?Â±Ã?Â¡. Suppose also that, in the plane Ã?Â±Ã?Â¡, a definite side of the straight line aÃ?Â¡ be assigned. Denote by hÃ?Â¡ a half-ray of the straight line aÃ?Â¡ emanating from a point OÃ?Â¡ of this line. Then in the plane Ã?Â±Ã?Â¡ there is one and only one half-ray kÃ?Â¡ such that the angle (h, k), or (k, h), is congruent to the angle (hÃ?Â¡, kÃ?Â¡) and at the same time all interior points of the angle (hÃ?Â¡, kÃ?Â¡) lie upon the given side of aÃ?Â¡. We express this relation by means of the notation Ã¢?Â (h, k) Ã¢?? (hÃ?Â¡, kÃ?Â¡)

Every angle is congruent to itself; that is, Ã¢?Â (h, k) Ã¢?? (h, k)

or Ã¢?Â (h, k) Ã¢?? (k, h)

IV.5: If the angle (h, k) is congruent to the angle (hÃ?Â¡, kÃ?Â¡) and to the angle (hÃ?Â¡Ã?Â¡, kÃ?Â¡Ã?Â¡), then the angle (hÃ?Â¡, kÃ?Â¡) is congruent to the angle (hÃ?Â¡Ã?Â¡, kÃ?Â¡Ã?Â¡); that is to say, if Ã¢?Â (h, k) Ã¢?? (hÃ?Â¡, kÃ?Â¡) and Ã¢?Â (h, k) Ã¢?? (hÃ?Â¡Ã?Â¡, kÃ?Â¡Ã?Â¡), then Ã¢?Â (hÃ?Â¡, kÃ?Â¡) Ã¢?? (hÃ?Â¡Ã?Â¡, kÃ?Â¡Ã?Â¡).

IV.6: If, in the two triangles ABC and AÃ?Â¡BÃ?Â¡CÃ?Â¡ the congruences AB Ã¢?? AÃ?Â¡BÃ?Â¡, AC Ã¢?? AÃ?Â¡CÃ?Â¡, Ã¢?Â BAC Ã¢?? Ã¢?Â BÃ?Â¡AÃ?Â¡CÃ?Â¡ hold, then the congruences Ã¢?Â ABC Ã¢?? Ã¢?Â AÃ?Â¡BÃ?Â¡CÃ?Â¡ and Ã¢?Â ACB Ã¢?? Ã¢?Â AÃ?Â¡CÃ?Â¡BÃ?Â¡ also hold.

V. Continuity

V.1: Axiom of Archimedes. Let A1 be any point upon a straight line between the arbitrarily chosen points A and B. Take the points A2, A3, A4, . . . so that A1 lies between A and A2, A2 between A1 and A3, A3 between A2 and A4 etc. Moreover, let the segments AA1, A1A2, A2A3, A3A4, . . . be equal to one another. Then, among this series of points, there always exists a certain point An such that B lies between A and An.

V.2: Line completeness. To a system of points, straight lines, and planes, it is impossible to add other elements in such a manner that the system thus generalized shall form a new geometry obeying all of the five groups of axioms. In other words, the elements of geometry form a system which is not susceptible of extension, if we regard the five groups of axioms as valid.

Lest I be criticized for citing Wikipedia, I provided the above directly from Townsend's transation, with some small modifications being made since from a German speaker who improved some translation points.

For about a year probably on the Wikipedia discussion page I've had a question about why I.4 is not implied by I.3. No takers.

Can any here advise?

There must be answer. Hilbert was no fool. There must be something I do not see.