The more difficult, complex, and better the exercises for congruent triangles, the higher the number

The problem of two overlapping regular polygons, one of which rotates around a vertex

Experiments and Demonstrations:

1) The degree of the solution is expressed by the formula containing : 3=60°- 4= , 5=36°-

2) In Fig. 1-4, when connecting A0H, is there a line segment that is perpendicular to the line A0H and bisected by it without adding other auxiliary lines? If it exists, please select one of the graphs to prove it; If not, please explain the reason;

Induction and conjecture:

Let the n-sided shape a0a1a2....an-1 with regular n-sided a0b1b2....bn-1 coincides (where a1 coincides with b1), and now the regular square a0b1b2....bn-1 rotates counterclockwise around vertex a0 (0° 0°n);

3) Let n be the same as the above "3, 4、...", please write the degree of n directly;

4) Guess whether there is a line segment perpendicular to the line a0h and bisected by it in the case of a regular n-sided shape? If it exists, please indicate the line segment with the corresponding vertex letter (proof is not required); If not, please explain the reason;

In trapezoidal ABCD, AD is parallel to BC, AB is equal to DC, and take a point F and G on AB and DC to make BF=CG, and E is the midpoint of AD.

Guarantee is difficult.

Draw a straight line from one of the vertices of any quadrilateral to divide the quadrilateral into two parts of the same area. (Congruent triangles are used).

Point E is outside ABC, point D is on the edge of BC, de is on the edge of BC, DE is on point F, if 1= state judgment 2= 3, AC=AE, try to say that the sail carries the change to the hidden rolling: the reason for ABC ADE.

Isosceles right triangle ABC, AB....

1.Do BM Vertical AC

en vertical df

ABM congruent den

Get bm=en

then mbc congruent nef

Get c= f

Get abc congruent def

2.Establish. Ditto for high.

Prove two congruences.

There is no SSA judgment axiom.

The two triangles are obtuse angles.

Or two triangles at the same acute angle.

It can be done high to obtain congruence.

It cannot be used directly.

Congruence doesn't seem to be provable.