Triangles ABC and ADC are on the same side of line AC. It is known that AB = CD, AD = CB

Triangles ABC and ADC are on the same side of line AC. It is known that AB = CD, AD = CB, M is the midpoint of AC. Prove that triangle BMD is isosceles.

Take a segment AC belonging to the line (AC). This line divides the plane into two half-planes. Let points B and D lie in the same half-plane. Let’s connect points B and D with points A and C. You get two triangles – ABC and ADC. By the condition of the problem:

| AB | = | CD |;

| AD | = | CB |;

Let us draw the segment BC further, and let the point M be the midpoint of the segment AC. It is required to prove that triangle BMD is isosceles.

For this:

prove the equality of triangles ABC and ADC;
show that BD || AC;
establish the equality of the segments MB and MD.
Equality of triangles ABC and ADC
As you know, if the sides of one triangle are pairwise equal to the sides of the other triangle, then these triangles themselves are also equal to each other. In our case, in relation to triangles ABC and ADC, we have:

| AB | = | CD |;

| AD | = | CB |;

by the condition of the problem, and the third party of the AU is common, i.e. the same in both triangles. This means that ∆ABC = ∆ADC, and the heights and medians drawn from the vertices B and D of these triangles to the AC side are also equal. Accordingly, points B and D are equidistant from the straight line (AC), and:

(BD) || (AC);

Note that the quadrilateral ABDC is an isosceles trapezoid.

BMD Triangle Properties
A triangle is isosceles if its two sides are equal. Note that BM is the median in ∆ABC. In triangle ADC, the segment DM is also the median. Respectively:

| BM | = | DM |;

We get that in ∆ВMD the sides BM and MD are equal. Therefore, ∆ВMD is isosceles, as required.