D is a point inside the triangle ABC such that DA+DB+DC is least.?

1) It is known as the Fermat Point. You can check it out on Wikipedia. It is a well know point.

2) Check my proof out. Let the sides of the triangle be a,b,c, usual notation, and without loss of generality assume a>b>c.

Now a(DX+DY+DZ) = aDX + aDY + aDZ >= aDX + bDY + cDZ = 2Ar(ABC)

For the last step we note that area of BCX is DX*a/2 and similarly for others and we add them up.

So a(DX+DY+DZ) >= 2Ar(ABC)

thus (DX+DY+DZ) >= 2Ar(ABC)/a = h1, where h1 is the height of the altitude from A.

Now u have given that D lies in the triangle. But if D can lie in the whole plane then clearly equality is given if D coincides with A, since DY and DZ become 0 and DX becomes the altitude itself. But even if D is inside the triangle, the value DX + DY + DZ can be made arbitrarily close to h1 by taking D close to A. So not exaclty the minimum value, but we can say that the infimum is h1.

the respond is 13.5 In words: The triangle BDE is a ingredient smaller than ABC. And all of us comprehend that BD is two contraptions long and BA is (2+7=) 9 contraptions long. which ability BD is 4.5 situations smaller than BA consequently DE could desire to be 4.5 situations smaller than AC. It follows then that AC is 3x4.5= 13.5 contraptions long. In symbols: enable u outline contraptions of length: BD = 2 u BA = 9 u BA/BD = AC/DE => 9/2 = AC/3 => AC = 3 x 4.5 = 13.5 u