Take point D in triangle ABC so that the sum of Da + DB + DC and the minimum D point is there Three companies form triangle lines in three locations of ABC, and the semi-finished products of the company come and go frequently. Therefore, we build a public warehouse location D. in order to optimize the freight of semi-finished products, we find the point d with the above conditions. We need to prove why the point D will have Da + DB + DC as the minimum

Take point D in triangle ABC so that the sum of Da + DB + DC and the minimum D point is there Three companies form triangle lines in three locations of ABC, and the semi-finished products of the company come and go frequently. Therefore, we build a public warehouse location D. in order to optimize the freight of semi-finished products, we find the point d with the above conditions. We need to prove why the point D will have Da + DB + DC as the minimum

Da = DB indicates that the vertical bisector of AB passes through point D
CD is the angular bisector of angle ABC
Angle BCD = 30 degrees
DB=DB
PB=AB=BC
Angle PBD = angle DBC
Triangle PBD is equal to triangle CBD
Angle BPD = angle BCD = 30 degrees

As shown in the figure, it is known that D is an internal point of △ ABC, and DB = DC, ∠ abd = ∠ ACD, and ab = AC

It is proved that: ∵ DB = DC, ∵ DBC = DCB, ∵ abd = ACD, ∵ abd + DBC = ACD + DCB, that is, ∵ ABC = ACB, ∵ AB = AC

As shown in the figure, it is known that D is an internal point of △ ABC, and DB = DC, ∠ abd = ∠ ACD, and ab = AC

It is proved that: ∵ DB = DC, ∵ DBC = DCB, ∵ abd = ACD, ∵ abd + DBC = ACD + DCB, that is, ∵ ABC = ACB, ∵ AB = AC

It is known that in the triangle ABC, ∠ ACB = 90 degrees, point D is on side AB, DB = DC, triangle ACD is isosceles triangle

∵DB=DC
∴∠DCB=∠B
And ∵ ∠ ACB = 90 °
∴∠B+∠A=90°
∴∠A+∠DCB=90°
And ∵ ∠ DCA + ∠ DCB = 90 °
∴∠A=∠DCA=90°
∴AD=DC
That is, ACD is an isosceles triangle

It is known that: as shown in the figure, in △ ABC, ab = AC, D is the point outside △ ABC, and ∠ abd = 60 ° and ∠ ACD = 60 ° prove: BD + DC = ab

It is proved that: extend BD to F, make BF = Ba, connect AF, CF, ∫ abd = 60 degrees, ∫ ABF is equilateral triangle, ∫ AF = AB = AC = BF, ∫ AFB = 60 degree, ∫ ACF = AFC, and ∫ ACD = 60 degree, ∫ AFB = ACD = 60 degree, ∫ DFC = DCF, ∫ DC = DF. ∫ BD + DC = BD + DF = BF = ab