Auxiliary points: turning a sum into a segment · Plane geometry I

A hypothesis like \(AB = AD + BC\) mentions three segments that live in three different corners of the figure. No congruence criterion can touch it while the sum is still a sum. The whole solution is one new point: the point that turns \(AD + BC\) into a single segment.

Turning a sum into a segment

Principle

When a length appears as a sum \(p + q\), either split: mark the point that cuts the long segment into a piece of length \(p\) and a piece of length \(q\); or extend: prolong the segment of length \(p\) beyond its endpoint by a piece of length \(q\). Both moves put \(p\) and \(q\) next to each other so that ordinary triangles can compare them.

Example

In a convex quadrilateral \(ABCD\) we are told that \(AB = AD + BC\). Mark \(E\) on \(AB\) with \(AE = AD\). Then automatically \(EB = AB - AE = BC\), and the single hypothesis has become two isosceles triangles, \(ADE\) and \(BCE\), sharing the vertex \(E\).

One tick: \(AE = AD\), by choice. Two ticks: \(EB = BC\), for free.

From there the angles do the rest: an isosceles triangle hands you an equal pair of base angles wherever you put it, and those angles are exactly what a conclusion about angles is asking for.

Reflections build the point for you

Three reflections cover most of what you will need:

In an angle bisector. It swaps the two arms of the angle, so it carries a point of one arm to a point of the other at the same distance from the vertex - this is the "cut the shorter side off along the longer one" move, done in one stroke.
In a line. Distances to points of the line are preserved, which is why \(PA + PB\) is minimised by reflecting one of \(A\), \(B\) across the line and taking the straight segment.
In a midpoint (a half turn). It carries a segment to a parallel segment of equal length, so it turns a median into a full parallelogram - see the next lesson.

Warning

Define the auxiliary point by exactly one property ("the point of \(AB\) with \(AE = AD\)"), then prove everything else about it. And check the configuration: \(E\) lies between \(A\) and \(B\) only because \(AD < AB\). A construction that silently assumes an order of points on a line is an incomplete proof.

Warm-up

Example

In a triangle \(ABC\) with \(\angle BAC = 60^\circ\), the bisectors \(BD\) and \(CE\) meet at \(I\) (with \(D\) on \(AC\) and \(E\) on \(AB\)). Prove that \(BE + CD = BC\).

What does the \(60^\circ\) actually control?

The angle between two internal bisectors: \(\angle BIC = 90^\circ + \tfrac{1}{2}\angle BAC = 120^\circ\). So the angles \(\angle BIE\) and \(\angle DIC\), which complete \(\angle BIC\) to a straight angle, are both \(60^\circ\).

The conclusion is a sum - split it

Mark \(F\) on \(BC\) with \(BF = BE\). It suffices to prove \(CF = CD\), and then \(BC = BF + FC = BE + CD\).

Two congruences pivoting on \(I\)

Triangles \(BIE\) and \(BIF\) share \(BI\), have \(BE = BF\) and equal angles at \(B\): congruent by SAS, so \(\angle BIF = \angle BIE = 60^\circ\). Hence \(\angle FIC = 120^\circ - 60^\circ = 60^\circ = \angle DIC\), and triangles \(CIF\), \(CID\) match by ASA (common side \(CI\), equal angles at \(C\), equal angles at \(I\)). Therefore \(CF = CD\).