Can anyone solve this geometry problem for junior high school?

The theorem that the opposite side of thirty degrees is half the hypotenuse cannot be proved in this way:

Connecting mb gives the triangle amn and bmn are congruent triangles (corner edge theorem) so am=bm so angle abm = 30 degrees i.e. mb is the angle bisector of the angle abc and because the angle acb = 90 degrees so mc = mn (the distance from the point on the angle bisector to the 2 sides is equal) and because the angle amn = 60 degrees so the angle nmc = 120 degrees so the angle acm = angle mnc = 30 degrees so the angle ncb = angle cnb = 60 degrees So the triangle BCN is an equilateral triangle (all 3 angles are 60 degrees).

The title is wrong, the perpendicular line of ab intersects ab, ac in m, n should be the perpendicular line of ac intersects ac, ab in m, n, if so, it is easy to prove: because the perpendicular line of ac intersects ac, ab in m, n, so cn=an, angle nca = angle a = 30 degrees, so angle bcn = 60 degrees, and in the right triangle abc, the angle acb is equal to 90 DEGREES, AND THE ANGLE A IS EQUAL TO 30 DEGREES, SO THE ANGLE B = 60 DEGREES, BECAUSE IN THE TRIANGLE BCN, THE ANGLE BCN = 60 degrees = angle b=60 degrees, so the triangle BCN is an equilateral triangle.

First of all, I have some doubts that you have written the question wrong, I think it should be "verify that the triangle BCM is an equilateral triangle", if it is correct, then I will not!!

First of all, the side of 30 degrees is half of the hypotenuse (in a right triangle), then the BC side is half of the AB side, and BM is also half of the AB side, BM=BC, and the B angle is 60 degrees, which proves that BCM is an equilateral triangle.

There is also a possibility to cross ab,ac to n,m

First of all, I would like to remind the landlord that AB is an hypotenuse, and the right-angled edge corresponding to the 30° angle in the right triangle is half of the hypotenuse (the specific landlord can prove it himself). Here's the process: ACB+ A=120° C=60° Mn is the perpendicular line of AB NB=1 2AB CB=1 2AB (in a right triangle, the right-angled side corresponding to a 30° angle is half of the hypotenuse) NB=CB BCN is an equilateral triangle (one angle is 60°, and the two sides are equal).

1.Dg is perpendicular to ab at point D

AD divides the bao

dao=∠gad

bao=∠dga=90°

ad=ad△doa≌△dga

oda=∠gda

OC Vertical AB

oca=∠dga=90°

gda=∠deo

oda=∠deo

oe=od2. ∵doa≌△dga

od=dg=oe=2

cos∠ofe=2/(2+

EF parallel AB

b=∠ofe

cos∠b=2/

bf=2ob=2+

The coordinates of b are (0,

Solution: EF parallel AB and OC AB

bco=∠oef=90°

and ad divided the bao

bad=∠oad

AB parallel EF

bad=∠def

fed+ ocd=90°

oad+∠oda=90°

and efd= dao

oda=∠oed

OD=OE Question 2 No.

If b is hypotenuse, from the Pythagorean theorem, c=4, at this time it is a right triangle, if c is an hypotenuse, from the Pythagorean theorem, it is obtained, c= 34, at this time it is a right triangle, so the triangle abc becomes an acute triangle c c is 4

Because a=3 b=5 if c=4 is a right triangle, then when 4

2. If it is to be an acute angle, the root number is under 34

Rotate ABP clockwise around point A so that AB coincides with AC, and turn point P to point D then AP=AD, so APD= ADP, and ADC= APB> APC so CDP> CPD

So the CD <>

(1) When BC=2AB, the quadrilateral PEMF is rectangular.

It is proved that BMC=90° is OK (pass M as the perpendicular line of BC, use the hypotenuse midline).

2) When D is the midpoint of BC, the rectangular PEMF becomes square (it is OK to prove that the adjacent sides are equal).

1 AD=2AB, because PEMF is the line of moments, so PF MC, 2 P is the midpoint of BC.