As shown in the figure, in RT △ ABC, ∠ C = 90 °, am is the middle line on the side of BC, sin ∠ cam = 35, then the value of tanb is () A. 32B. 23C. 56D. 43

As shown in the figure, in RT △ ABC, ∠ C = 90 °, am is the middle line on the side of BC, sin ∠ cam = 35, then the value of tanb is () A. 32B. 23C. 56D. 43

In RT △ ACM, sin ∠ cam = CMAM = 35, let cm = 3x, then am = 5x. According to Pythagorean theorem, AC = AM2 − cm2 = 4x, and M is the midpoint of BC, ﹥ BC = 2cm = 6x, in RT △ ABC, tanb = ACBC = 4x6x = 23

As shown in the figure, △ ABC and △ CDE are isosceles right triangles, points B, C and D are on a straight line, and point m is the midpoint of AE. The following conclusions are drawn: ① Tan ∠ AEC = BCCD; ② s △ ABC + s △ CDE ≥ s △ ace; ③ BM ⊥ DM; ④ BM = DM. The number of correct conclusions is ()
A. 1 B. 2 C. 3 d. 4

The ABC and CDE are all right triatriatriatriangles, and ABC + s △ CDE = 12 (A2 + B2) ≥ AB (take the equal sign when a = b), s △ ABC + s △ CDE ≥ s △ ace; so this option is correct; ④ through point m, Mn is perpendicular to BD, and the perpendicular foot is n. ∵ point m is the midpoint of AE, Mn is the trapezoidal median line, ∵ n is the midpoint, ∵ BMD is the isosceles triangle, ∵ BM = DM; so this option is correct; ③ and Mn = 12 (AB + ed) = 12 (BC + CD), ∵ BMD = 90 °, that is BM ⊥ DM; so this option is correct So D

If AB = AC = 10 in △ ABC is known, de bisects AB vertically and intersects AC with E. if the perimeter of △ BEC is known to be 16, find the perimeter of △ ABC

∵ de vertically bisects AB, ∵ AE = be, ∵ CE + be = CE + AE = AC, and the perimeter of △ BEC is 16, ∵ AC + BC = 16 ∵ BC = 16-10 = 6 ∵ ABC is BC + AC + AB = 10 + 10 + 6 = 26

In the triangle ABC, ab = 10, de bisects AB vertically, AC intersects e, perimeter of triangle BEC = 16, perimeter of triangle ABC=

∵ e is on the vertical bisector of ab
∴EA=EB
The circumference of ∧ BCE - BC + CE + be = AC + BC
∵△ BCE perimeter = 16
∴AC+BC=16
The circumference of ABC = 16 + AB = 16 + 10 = 26

In triangle ABC, de bisects AB vertically, the perimeter of triangle BEC is 20, BC = 9, and the perimeter of triangle ABC is calculated
In the triangle ABC, ab = AC, ∠ a = 36 ', de bisects AB vertically, the perimeter of the triangle BEC is 20, BC = 9, finding the perimeter of the triangle ABC is only so much

The perimeter of triangle BEC = be + EC + BC = 20, BC = 9, be + EC = 11, ab = AC
The perimeter of triangle ABC = 11 + 11 + 9 = 31

As shown in the figure, in diamond ABCD, ∠ a = 110 °, e and F are the midpoint of edge AB and BC respectively, EP ⊥ CD is at point P, then ∠ FPC =?

In the diamond ABCD, e and F are the middle points of edges AB and BC respectively, so be = BF ∠ bef = ∠ BFE = 55 degrees, (according to the inner angle and 180 degrees of the triangle), take the middle point I of AD, connect FI and intersect EP at o point, because BF = FC, so EO = OP, so EF = FP, so ∠ FEP = ∠ FPE, so ∠ FPC = 90 - ∠ FPE =

As shown in the figure, in the diamond ABCD, if the angle a = 110 degrees, e and F are the midpoint of AB and BC respectively, and EP is perpendicular to CD and P, then FPC = - (to process,

It can be proved that △ BGF ≌ △ CPF ≌ f is the midpoint of PG, and ∵ from the title, it can be seen that ≌ △ BEP is 90 °

As shown in the figure, in diamond ABCD, ∠ a = 110 °, e and F are the midpoint of edges AB and BC respectively, and EP ⊥ CD is at point P. calculate the degree of ∠ FPC

In diamond ABCD, ∵ a = 110 °, ∵ B = 180 ° - 110 ° = 70 °, ∵ e, f are the midpoint of edge AB and BC respectively, ∵ be = BF, ∵ bef = 12 (180 ° - b) = 12 (180 ° - 70 °) = 55 °, ∵ EP ⊥ CD, ab ∥ CD, ∵ BEP = ∠ CPE = 90 °, ∵ FEP = 90 ° - 55 ° = 35 °, take a

As shown in the figure, in diamond ABCD, ∠ a = 110 °, e and F are the midpoint of edges AB and BC respectively, and EP ⊥ CD is at point P. calculate the degree of ∠ FPC

In diamond ABCD, ∵ a = 110 °, ∵ B = 180 ° - 110 ° = 70 °, ∵ e, f are the midpoint of edge AB and BC respectively, ∵ be = BF, ∵ bef = 12 (180 ° - b) = 12 (180 ° - 70 °) = 55 °, ∵ EP ⊥ CD, ab ∥ CD, ∵ BEP = ∠ CPE = 90 °, ∵ FEP = 90 ° - 55 ° = 35 °, take the midpoint g of AD, connect FG with EP at O, ∵ point F is the midpoint of BC, ∵ FG ∥ CD, ⊥ EP ⊥ CD, ≁ EP ⊥ FG G vertical bisection of EP, ∧ EF = PF, ∧ FPE = ∧ FEP = 35 °, ∧ FPC = 90 ° - ∧ FPE = 90 ° - 35 ° = 55 °

In the diamond ABCD, angle a = 110 degrees, e and F are the midpoint of edge AB and BC respectively, EP is perpendicular to point P, and angle FPC = []

Let the side length of the diamond be a
EP=asin70°
EF=asin35°
FP²=（asin70°）²+（asin35°）²-2a²sin70°sin35°cos35°=(asin35°）²
FP=asin35°=EF
The triangle EFP is isosceles triangle, ∠ FEP = = FPE = 35 degree
∠FPC=90°-35°=55°