A middle school geometry proof problem that stumped many college students and asked the master to so

Is it really that hard? I'll try:

Extend CD to G so that DG = Be, linking AGApparently rt abe rt adg, so there is ae=ag , bae= dagand bae+ fad=90°- eaf=45°, thus giving gaf= dag+ fad=45°= eaf

In AEF and AGF, AEF AGF is obtained by , and the common edge af=af, so ah=ad (the height of the corresponding side of the congruent triangle is equal).

It is also possible to obtain AFE= AFD (i.e., AFG) after obtaining AEF AGF, and from ahf= ADF=RT (i.e., right angles), we know that the quadrilateral AHFD is a circle with a quadrilateral, so ah=AD (in the same circle or equal circle, the strings opposite to the equal central angles of the circle are also equal).

Note: 1. RT is a right triangle.

2. The auxiliary line method of this proof is a commonly used figure flipping method in geometric proofs, that is, the essence is to flip AEF along the straight line AF to the position of AGF. It can also be said to be the rotation method, which is to rotate the ABE around the point A to the position of the ADG.

It seems that you don't have to be a college student to prove it, hehe. If you don't give 5 points, you can't do it.

After A as AG perpendicular to AE CD extension line at G, the triangle AGD is fully equal to the triangle AEB, the triangle AGF is equal to the triangle AEF, and the triangle AGD is equal to the triangle AEH, and AD=AH is obtained

This can be demonstrated by a special case, where AH is on the diagonal AC.

It is easy to prove that the Rt triangle AME and PQE are congruent with the BC perpendicular line over Q and the BC perpendicular line is crossed by F, so that ME=EQ; In a square ABCD, the triangle AFD is congruent with CFD, thus, AF=FC, according to the nature of the perpendicular bisector, AF=FP, i.e., FP=FC, in the isosceles triangle PFC, the high FR on the bottom edge bisects PC, thus in the right-angled trapezoidal PQNC, FQ=NF, so EF=EQ+QF=ME+NF

This is normal, because there is a course called "Elementary Geometry" in college, which is to study this aspect, even the high school Olympiad involves the content of plane geometry, so plane geometry is broad and profound.

The title provided by the landlord is not very difficult.

Extending Ca to F, connecting Fe, Fb makes it easy to prove that the triangle CEBF is a parallelogram row, so CE Fb, so CFB= ECA=90°

Fe BC again

So feb= ebc

deb=∠dfb=90°

DBFE is a four-point circle.

feb= cdb (equal circumferential angle of the same arc) ebc= cdb

In addition, it is not necessarily good for college students to solve problems. Although I also became a teacher a year after graduating from college, I am very fortunate to have some indescribable feelings.

ae：ad=ac：ae

ae=abab：ad=ac：ab

The angles between the two sides are DAB

So cab is similar to bad.

Connect to CQ PQ

Then bpq is also an equilateral triangle bpq=60° in pqc, pq=pb= 3 qc=2 pc=1, so qpc=90°

So BPC= BPQ+ QPC=150°

Have you ever studied the cosine theorem?

Are the conditions finished ?? There should still be conditions.

Taking the midpoint n of BC and connecting NF and AN, considering the median line theorem of the triangle and the properties of isosceles triangles, it will definitely be proved.

Solution: Connect EF

Point E is the midpoint of AB and point F is the midpoint of AC.

EF is the median line of ABC.

EF = 1 2BC, EF BC (the median line of the triangle is parallel to the third side and equal to half of the third side).

Big brother, do you understand the median line of EF?