How to prove that the segment $FG= DJ+EH$

Question

$BE, CD, AF$ are the angle bisectors of the $\triangle ABC$. A circle passes through the points $D, E, F$ and intersects the triangle again at $J, H, G$. How to prove that $FG=DJ+EH$?

I tried to use angle bisector theorem with intersecting secant theorem and got some equations. Am I on the right track? Any hints or answers?

Answer 1

Let the angle bisectors of $\triangle ABC$ meet the opposite sides at $D$, $E$, $F$, and let $\bigcirc DEF$ meet the sides again at $D'$, $E'$, $F'$. Define signed lengths $d:=|DD'|$, $e:=|EE'|$, $f:=|FF'|$, so that the task is to show $d+e+f=0$.

By the Intersecting Secants Theorem, using signed distances oriented with the vertex order $ABC$, we have $$\begin{align} |AF||AF'| = |EA||E'A| &\quad\to\quad |AF|(|AF|+|FF'|)=|EA|(|E'E|+|EA|) \\ &\quad\to\quad |AF|(|AF|+f)=|EA|(-e+|EA|) \tag{1} \end{align}$$ By the Angle Bisector Theorem, $$|AF| = |AB|\cdot \frac{|CA|}{|BC|+|CA|}=\frac{bc}{a+b} \qquad\qquad |EA| = \frac{bc}{c+a} \tag{2}$$ so that $(1)$ becomes $$(a + b)^2 (a + c) e +(a + b) (a + c)^2 f = b c (b - c) (2 a + b + c) \tag{3}$$ Likewise, $$\begin{align} (b + c)^2 (b + a) f +(b + c) (b + a)^2 d &= c a (c - a) (a + 2b + c) \\ (c + a)^2 (c + b) d +(c + a) (c + b)^2 e &= a b (a - b) (a + b + 2c) \end{align}\tag{4}$$ Solving yields $$(d,e,f) = \frac{ \left(\; (b - c) (a b c - p^3 + 2 a p^2), (c - a) (a b c - p^3 + 2 b p^2), (a - b) (a b c - p^3 + 2 c p^2) \;\right)}{2 (a + b) (b + c) (c + a)}\tag{5}$$ where $p:=a+b+c$. From here, we readily see that $d+e+f=0$, as desired. $\square$

Answer 2

This is just a suggestion and there must be some more eaiser and elegant ways.

Fact#1 Length of chord is equal to 2R sin x; where R is the radius of the respective circle and x is angle subtended by the chord at the circumference.

Fact#2 $\sin a + \sin b = 2 \sin (0.5(a + b)) \cos (0.5(a – b))$

To prove the required result, we need $2R \sin (large) = 2R \sin (medium) + 2R \sin (small)$. Which, in turn, is $\sin (large) = \sin (medium) + \sin (small)$. The RHS matches the LHS of fact #2.

The 0.5(a + b) and the 0.5(a – b) let me think of the angles formed by the near arc and the far arc.

Hope those can help. Good luck.