Question

Point D is the in center of triangle ABC. Write an expression for the length x in terms of the three side lengths AB, AC, BC.

Answer 1

Answer :

$\underline{ \boxed{x = \frac{2(A_T) }{(AB) + (AC) + (BC)} }}$

Explanation :

$i \: figured \: it \: out \to \: notice \: that \\ there \: are \:( thre e\: trianges) \: in \: triangle \: ( ABC) \\ and \: they \: all \: have \: the \: same \: height \: = x \\ first \: triangle \: is \: \to \: triangle \: (ABD) \\ second\: triangle \: is \: \to \: triangle \: (ACD) \\ third \: triangle \: is \: \to \: triangle \: (BCD) \\ if \: the \: area \: of \: a \: triangle \: is \: = \frac{1}{2} bh : then \to \\ the \: area \: of \: triangle \: (ABD) = \frac{1}{2} (AB)x \\ the \: area \: of \: triangle \: (ACD) = \frac{1}{2} (AC)x \\ the \: area \: of \: triangle \: (BCD) = \frac{1}{2} (BC)x \\ let \: the \: total \: area \:(A_T) \: be \to \: \frac{1}{2} (AB)x + \frac{1}{2} (AC)x + \frac{1}{2} (BC)x \\ if \to\\ (A_T) = \frac{1}{2} (AB)x + \frac{1}{2} (AC)x + \frac{1}{2} (BC)x \\ then \to \\ (A_T) = \frac{1}{2} \{(AB)x + (AC)x + (BC)x \} \\ x \{(AB) + (AC) + (BC) \} = 2(A_T) \\ x = \frac{2(A_T) }{(AB) + (AC) + (BC)}$