In the triangle ABC, the angle BAC = 100 degrees, de and FG bisect AB and AC vertically and intersect BC at e and G, respectively

In the triangle ABC, the angle BAC = 100 degrees, de and FG bisect AB and AC vertically and intersect BC at e and G, respectively

Right triangle BDE is equal to right triangle ade, so angle B = angle EAD;
Similarly, angle c = angle GAF
The sum of internal angles of triangle ABC is 180, so the sum of angles B and C is 180-110 = 70
The sum of angle ead and angle GAF is 70, so angle EAG = 110-70 = 40 degrees

As shown in the figure, ∠ BAC = 110 ° in △ ABC, e and G are the midpoint of AB and AC respectively, de ⊥ AB and FG ⊥ AC, and ∠ DAF is calculated

∵⊥ BAC = 110 °, ∵ B + ⊥ C = 180 ° - 110 ° = 70 °, ∵ E and G are the midpoint of AB and AC respectively, de ⊥ AB, FG ⊥ AC, ∵ ad = BD, AF = CF, ∵ bad = - B, ∵ CAF = - C, ∵ DAF = - BAC - (≁ bad + ⊥ CAF) = - BAC - (≁ B + ⊥ C) = 110 ° - 70 ° = 40 °

Let G be the center of gravity of △ ABC, Ag = 6, BG = 8, CG = 10, then the area of △ ABC is

Because G is the center of gravity of the triangle ABC,
So GA + GB + GC = 0 (vector),
From (- GA) ^ 2 = (GB + GC) ^ 2 = GB ^ 2 + GC ^ 2 + 2GB * GC
36=64+100+2GB*GC ,
The solution is GB * GC = - 64,
Similarly, GA * GB = 0, GA * GC = - 36,
So cos ∠ BGC = GB * GC / (| GB | * | GC |) = - 4 / 5,
Similarly, cos ∠ AGB = 0, cos ∠ CGA = - 3 / 5,
So sin ∠ BGC = 3 / 5, sin ∠ AGB = 1, sin ∠ CGA = 4 / 5,
Therefore, SABC = SAGB + SBGC + SCGA
=1/2*|GA|*|GB|*sin∠AGB+1/2*|GB|*|GC|*sin∠BGC+1/2*|GC|*|GA|*sin∠CGA
=1/2*6*8*1+1/2*8*10*3/5+1/2*6*10*4/5
=72 .
(in fact, the lengths of the three midlines are 9, 12 and 15 respectively, and a triangle is similar to the triangle formed by its three midlines, and the area ratio is 4:3)