/posts/00/32/a3/16/a3318550.jpg is known as shown in the figure: △ ABC, ∠ 1 = ∠ 2, ∠ 3=∠ 4,. Verification: AB = AC [B] Analysis: Comparing the lengths of two line segments, there are only three cases. If AB is not equal to AC, then there are only two situations, either AB > AC, or AB can disprove it, and it can only be the third answer, that is, AB = AC, as long as the above two-clock hypothesis is proved to be untenable. (law of excluded middle in contradiction method, negation of negation) [/B] Proof: Let EH // BF, EH = BF, and connect FH and HC to form ∠ 5, ∠ 6, ∠7. There are ∠ 1+∠ 2 =∠ ABC, ∠ 3+∠ 4 = ∠ ACB, ∠ 4+∠ 7 = ∠ ECH, ∠5+∞. ∠ 1 = ∠ 6，HF =EB。 (1) If AB > AC, ∠ △ABC and ∠ 6 = ∠ 1, the diagonal of the parallelogram is equal to ∠ 6, so ∠ 7 △. BE compared in two △BCE and △BCF, △: because the two quantities are equal (BC = CB, BF = CE)△: By BE △: So ∠ ABC > ∠ ACB (double-angle equivalence relation) △: Therefore: AB Therefore: this result is the same as the hypothetical condition: in △ABC. (2) In the second case of △ABC, suppose AB is also provable; The conclusion that AB > AC is contradictory to the hypothesis that AC > AB in △ABC is contradictory, so the above hypothesis (2) cannot be established either! Because AB is not equal to AC, there are only the above two cases, but neither of them can be established, so only one case can be established, that is, AB = AC△△ proves that this is over-. It is proved that let D and E be the midpoint of BC and C'A', let AD and B'E intersect at G, connect ED and extend AB to F ... straight EDF section △A'BD=CD', which is obtained by Menelaus theorem: (C 'e/EA') * (A'd/DB) * (BF/FC'). Therefore, A'C/BD=C'B/BF. According to the known conditions: CA'/BC=BC'/AB, BD/BC=BF/AB= 1/2. So DF=CA/2, DF‖CA, DE = AB'/2, DE. Complete the certificate. It is known that in triangle ABC, be and cf are bisectors of angles, and d is the midpoint of EF. If d reaches the three sides BC, AB and AC of the triangle, then point E is the intersection of AB and BC at M and N, and point F is the intersection of AC and BC at P and Q. According to the equal distance between points on the bisector, FQ = FP and EM = EN. Point D is the intersection of BC at point O, point D is the intersection of AB at point H and point D is the intersection of AB at point H and because FQ = FP, EM = en. FQ = 2dj, um = 2hd. Because the angles FQC, DOC and ENC are all 90 degrees, the quadrilateral FQNE is a right trapezoid and d is the midpoint, so 2do = fq+en, fq = 2dj and en = 2hd. So do = HD+JD。 Because x = do, y = hy and z = DJ, x = y+z.-In trapezoidal ABCD, AB//CD, AD is perpendicular to AC, AD = AC, DB = DC, AC. BGD angle equals ACD angle equals 45 degrees. Let AD=AC= 1. Then DC=BD= root number 2, BG= 1. According to sine theorem, the sine value of BGD of BD edge is equal to that of BDG edge. Let the angle BDG= angle A, that is, the root number 2 is 45 = 1 compared with the previous Sina, so Sina equals half, so the angle BDG equals 30. As shown in the figure: 1 In △ABC, d is the midpoint of BC and DE is perpendicular to DF. Try to judge the relationship between BF+CF and EF, and try to prove the conclusion. 2 △ABC is an isosceles right triangle, ∠ACB=90 degrees, D is the midpoint of AC, connecting BD is ∠ADF=∠CDB, connecting BD in CF and E, and supplementing the problem that BD is perpendicular to CF: The solution of the first problem is as follows: BE+CF & gt；; EF extends ED, making DG=DE, connecting CG and FG, and triangle DEB is equal to triangle GCD, so BE=CG because DE=DG, DF=DF, angle EFD= angle FDG=90 degrees, so FG=EF because CF+DG & gt；; FG (the sum of two sides is greater than the third side) GF=BE, FG=EF, so BE+CF & gt；; The solution of EF's second problem is as follows: establish a plane rectangular coordinate system with point C as the far point, CB as the Y axis and CA as the X axis. Assuming that CA=CB=6, D (3 3,0) can be obtained; The analytical formula of B (0 0,6)-straight line DB is f (db) =-2x+6; The intersection point F is the vertical AC of FG, because ∠ DAF = 45, so AG = FG is from △DCB, and it can also be deduced that FG=2GD and AD=3, so FG = 2 and GD = 1, so F (4 4,2); The analytical formula of D (3 3,0)-straight FD is f (FD) = x/2; Because the product of k values of two analytic expressions is equal to-1, it can be proved that BD ⊥ CF.-.