# Prove that a circle can be inscribed iff the given condition is satisfied

I have the following question with me:

"Let $$B_1$$ and $$C_1$$ be points on the sides $$AC$$ and $$AB$$ of a triangle $$ABC$$. Lines $$BB_1$$ and $$CC_1$$ intersect at point $$D$$. Prove that a circle can be inscribed inside quadrilateral $$AB_1DC_1$$ if and only if the incircles of $$ABD$$ and $$ACD$$ are tangent to each other." I start like this:

I introduce notations:

$$AB = c, AC = b, BC = a, AC_1 = x, AB_1 = z, B_1D = y, C_1D = w, BD = q, CD = p$$

To prove the above statement I basically need to prove that the statements $$x + y = z + w$$ and $$c + p = b + q$$

are equivalent. Performing a few computations involving bases I get the following equations :

$$qz(p + w) = bw(q + y)$$ and $$yc(p + w) = px(q + y)$$

How do I proceed from here? Any help is appreciated. Alternative solutions are also welcome

This is actually a problem from Bulgarian MO, third round, 1999.

For your reference I have copied the solution from "Cyclic quadrilaterals, radical axis and inscribed circles" by Charles Leytem

For variety , here is a different solution. Let us suppose that the incircles of $$\triangle$$s $$ABD$$ and $$ACD$$ are not tangential . Clearly , as in the figure , there are two distinct points of tangency , $$G$$ and $$K$$

Let $$AG$$ = $$x$$ , $$GK$$ = $$\delta$$ and $$KD$$ = $$y$$ . The line segments of the same color in the figure are equal. Now consider quadrilaterals $$ABDC$$ , $$AB_1DC_1$$ . Let us assume that a circle can be circumscribed by $$AB_1DC_1$$ . Join points of tangency to form segments $$K’L’$$ , $$M’N’$$ . Join $$JL$$ , $$FH$$ . The respective dotted segments are parallel , by Thales’ Theorem.

Let $$DL’ = a$$ , $$L’B_1 = b$$ , $$M’A = c$$ and $$K’C_1 = d$$ . Let $$C_1J = q_1$$ and $$B_1H = q_2$$.

Using properties of tangents and isosceles $$\triangle$$s, we have:- $$AJ = x + \delta = c + d - q_1$$ $$AH = x = c + b - q_2$$ From these , we get $$\delta = d - b - q_1 + q_2$$ Call this equation (1). Also , we have :- $$LL’ = a + y = d - q_1$$ $$FN’= a+ \delta + y = b - q_2$$ From these , we get :- $$\delta = b-d+q_1-q_2$$ Call this equation (2) . Clearly , from (1) and (2) , we get $$\delta = 0$$ , a contradiction if the points are distinct . QED

Note:- This is a case in which $$H$$ and $$J$$ lie inside the quadrilateral $$AB_1DC_1$$ , while $$F$$ and $$L$$ lie outside it . There exist other cases , which can be proven similarly