1. As shown in figure 1, in acute angle △ABC, CD is perpendicular to AB and at point D, and E is a point on AB. Find all the acute triangles in the picture and explain the reasons. See figure:

3. As shown in Figure 2, in △ABC, ∠B is equal to ∠ c and ad is equal to ∠BAC, which shows why ∠ADB-∠ADC=∠C-∠B is established. See figure:

3. As shown in Figure 3, it is known that BO bisects ∠CBA, CO bisects ∠ ACB, Mn ‖ BC, AB = 12, AC = 18, and the circumference of △AMN is found. See figure:

4. As shown in Figure 4, it is known that in △ABC, AD is the high line on the side of BC and AE is the bisector of ∠BAC. If ∠EAD=a, find ∠ C-∠ B. (expressed by the algebraic expression of A), as shown in the figure:

5. As shown in Figure 5, it is known that AB=AC, AD=AE, ∠ 1=∠2. CE=BD? Explain why. See figure:

6. As shown in Figure 6, make equilateral triangles BCE and CDF from the sides BC and CD of the square ABCD, and connect AE, AF and EF, which proves that △AEF is equilateral triangle. See figure:

Question 1: There are six triangles in the figure 1 * *, namely △ABC, △AEC, △CED, △CBD, △ACD and △ECB, where △ CED, △ ACD and △ CDB are RT △△△ AEC is obtuse △. 90 △ABC acute angle △, known condition. ∞∠CEB = 180- obtuse angle = acute angle ∠B is acute angle, ∠ECB=∠ACB-∠ACE = acute angle △ECB is acute angle △ * * There are two acute angles△, and Question 2:∫ad is ∠. = AB+AC =12+18 = 30 ÷ circumference of AMN = 30 Question 4 ∠ C = 90-∠ DAC = 90-[(1/2) ∠ BAC. ∠ BAC ∠ C-∠ B = 90-[(1/2) ∠ BAC-a]-{90-a-(1/2) ∠ BAC} = 2A Question 6 ∠. The length of the equilateral triangle ABC is a, so take a point D on the extension line of BC to make △ECD=b, and a point E on the extension line of BA to make AE = A.. .