A math and geometry problem to challenge your brain power

Choices of 4,123 prove that the two triangles are equal. If you want to know the process in detail, please ask, thank you!

3. There is a case of inaccuraccy.

It's all sss, and 4 can only be similar.

I always thought that my eyesight was good, but this time, there were two sentences that I didn't see clearly, so please type them out.

Child: It's not that I can't do it, but I can't read the handwriting clearly.

1=120°, 2=105°, cf ab de (known) acf = 180°- 1 = 60° (two straight lines are parallel, complementary to the side and inner angles) The same is true: dcf = 180° - 2 = 75°

acd=∠acf+∠dcf=135°

3+ ACD = 180° (180° flat angle) 3 = 180°- ACD = 45°

2.Set it to be the right x.

5x-1（20-x）=76

x=163.Set up x trees.

3x+5=5（x-1）

x=53*5+5=20 (only).

Even be, then abedc is pentagonal, and the sum of internal angles is 540° ab de;

b+∠c=180°

again 1=120°, 2=105°

ACD=540°-180°-120°-105°=135° 3 is the complementary angle of ACB.

Proof: as dq bf, the intersection fb point is q

As dp ce, the intersection point is p

DCE and DBF are equal in area.

That is, 1 2dq*bf=1 2dp*ce

bf=ce

dq=dpdq⊥bf，dp⊥ce

dqb=∠dpc=90°

The perpendicular line of ab and ac is made at the point d, because the area of the two small triangles is equal, so the height of the two new works is equal, and the distance between the two sides of the point to the angle is equal, indicating that the point is on the bisector of this angle.

The area of the triangle is equal to 1 2 base * height, because, the area of DCE and DBF is equal, CE=BF, therefore, the height of CE and BF is equal.

That is, with d as the vertex, the high DM and DN are equal to both sides of CE and BF.

So AD bisects BAC (a characteristic of the angular bisector or then a congruence proof, the right triangle HL theorem).

Proof: Make perpendicular lines from point D to BF and CE, respectively DM and DN because the areas of DCE and DBF are equal.

So dm=dn

Because the distance from the point on the angular divider to both sides of the corner is equal.

So AD divides the bac equally

First question:

1. Connect to the AC

2. Set half of the angle DCB to be X, and half of the angle DAB to Y

3. According to the relationship between the triangle DOA and BOC (the inner angle of the triangle and 180 degrees), it is obtained: b+2x=d+2y, and x-y=(d-b) 2 equation 1 is introduced

4. According to the relationship between the triangle AEC and AOC and BOC (or DOA), (the inner angle of the triangle and 180 degrees), it is obtained: e=180-(x+y)-(180-b-2x)=(x-y)+b Eq. 2

5. Substituting Eq. 1 into Eq. 2 gives E=(B+D) 2 Eq. 3 (the answer to the first question).

The second question: 1. According to the given conditions of the question, it is obtained: d=2b, e=bx 2

2. Substitute the above into Equation 3

Derives: bx 2=3b 2 calculates, and gives x=3 answer to the second question.

This problem is relatively simple, because you only need to make an auxiliary line, repeatedly use the internal angle and theorem, and combine it with a little algebraic calculation.

However, it is more a test of patience, or hard work, so students who can't solve it should pay attention to using more scratch paper and typing more drafts, just looking at it is not enough.

I wish you all the best every day!