In △ ABC, ∠ a = 40 °, the bisectors BD and CE of ∠ B and ∠ C intersect at m, then ∠ BMC

In △ ABC, ∠ a = 40 °, the bisectors BD and CE of ∠ B and ∠ C intersect at m, then ∠ BMC

∵∠A=40°
∴∠B+∠C=180-40=140°
BD and CE are bisectors
∴∠MBC+∠MCB=1/2X(∠B+∠C)=70°
∴∠BMC=180-70=110°

As shown in the figure, in △ ABC, point D is on AC and point E is on the extension line of BC

It is proved that: ∵ DCB is an outer angle of △ DCE (definition of outer angle) ∵ DCB ＞ CDE (an outer angle of a triangle is greater than any inner angle not adjacent to it) ∵ ADB is an outer angle of △ BCD (definition of outer angle) ∵ ADB ＞ DCB (an outer angle of a triangle is greater than any inner angle not adjacent to it) ∵ ADB ＞ CDE (property of inequality)

As shown in the figure, the middle line of △ ABC is ad and be intersects at point F. if the area of △ ABC is 45, calculate the area of quadrilateral dcef

The midline connecting de and ∫ ABC is ad, be, ∫ de = 12ab, de ∥ AB, ∫ CDE ∫ CBA, the area of ∫ ABC is 45, ∫ 45s ∫ CDE = 4, ∫ s ∫ CDE = 11.25, ∫ de ∥ AB, ∫ def ∫ ABF, ∫ deab = efbf = dfaf = 12, the area of ∫ ABC is 45, be is the midline of ∥

As shown in the figure, let △ ABC and △ CDE be regular triangles, and ∠ EBD = 62 °, then the degree of ∠ AEB is ()
A. 124°B. 122°C. 120°D. 118°

∵ △ ABC and △ CDE are equilateral triangles, ∵ AC = BC, CE = CD, ≌ △ ACB = ≌ ECD = 60 °, ACB = ≌ ACE + ≌ BCD, ≌ DBC = ≌ CAE, i.e. 62 ° - EBC = 60 ° - BAE, i.e. 62 ° - (- 60 ° - Abe) = 60 ° - BAE