Let $t$ meet $AB$ and $BC$ at $P$ and $Q$, respectively. Let $PQ$ meet the line $AC$ at $Z$. We have to show $BX \cdot BY= AX \cdot AY+ CX \cdot CY$ $BX \cdot BY \cdot sin XBY= AX \cdot AY \cdot sin XAY + CX \cdot CY \cdot sin XCY$ $2[BXY]=2[AXY]+2[CXY]$ $1=\frac{[AXY]}{[BXY]}+ \frac{[CXY]}{[BXY]}$ $1=\frac{AP}{PB}+ \frac{CQ}{QB}$ By Menelaus, $\frac{ZT}{ZA} \cdot \frac{AP}{PB} \cdot 2=1$ and $\frac{ZC}{ZT} \cdot \frac{1}{2} \cdot \frac{BQ}{QC}=1$ Thus we find $\frac{AP}{PB}= \frac{ZA}{2ZT}$ and $\frac{CQ}{QB}=\frac{ZC}{2ZT}$ So $\frac{AP}{PB}+ \frac{CQ}{QB}= \frac{ZA}{2ZT}+\frac{ZC}{2ZT}$, and we have to show this equals $1$, or $ZA+ZC=2ZT$. But $ZA+ZC=(ZA-AT)+(ZC+CT)=2ZT$, which completes the proof.

Let $ N$ be the midpoint of $ AC$ and $ A',B',C',N'$ the orthogonal projections of $ A,B,C,N$ onto line $t.$ Segment $ NN'$ becomes the median of the trapezoid $ ACC'A',$ thus $ NN' = \frac {_1}{^2}(AA' + CC').$ But from $ \triangle MBB' \sim \triangle MNN'$ we get $ \frac {BB'}{NN'} = \frac {BM}{NM} = 2 \Longrightarrow \ BB' = AA' + CC' \ (\star)$ Let $R$ be the circumradius of $\triangle ABC.$ Using that the altitudes and circumdiameters issuing from the same vertices are isogonals, we obtain $BX \cdot BY = 2R \cdot BB' \ , \ AX \cdot AY = 2R \cdot AA' \ , \ CX \cdot CY = 2R \cdot CC'$ Combining these latter expressions with $ (\star)$ yields $ \frac {BX \cdot BY}{2R} = \frac {AX \cdot AY}{2R} + \frac {CX \cdot CY}{2R} \Longrightarrow BX \cdot BY = AX \cdot AY + CX \cdot CY.$

let $XY$ intersect $AB,BC$ at $S,T$ respectively. then $\frac{BX*BY}{CX*CY}=\frac{BT}{TC},\frac{BX*BY}{AX*AY}=\frac{BS}{SA}$ it suffices to prove $\frac{AS}{SB}+\frac{CT}{TB}=1$ which is trivial

dgreenb801 wrote: Let $t$ meet $AB$ and $BC$ at $P$ and $Q$, respectively. Let $PQ$ meet the line $AC$ at $Z$. We have to show $BX \cdot BY= AX \cdot AY+ CX \cdot CY$ $BX \cdot BY \cdot sin XBY= AX \cdot AY \cdot sin XAY + CX \cdot CY \cdot sin XCY$ $2[BXY]=2[AXY]+2[CXY]$ $1=\frac{[AXY]}{[BXY]}+ \frac{[CXY]}{[BXY]}$ $1=\frac{AP}{PB}+ \frac{CQ}{QB}$ By Menelaus, $\frac{ZT}{ZA} \cdot \frac{AP}{PB} \cdot 2=1$ and $\frac{ZC}{ZT} \cdot \frac{1}{2} \cdot \frac{BQ}{QC}=1$ Thus we find $\frac{AP}{PB}= \frac{ZA}{2ZT}$ and $\frac{CQ}{QB}=\frac{ZC}{2ZT}$ So $\frac{AP}{PB}+ \frac{CQ}{QB}= \frac{ZA}{2ZT}+\frac{ZC}{2ZT}$, and we have to show this equals $1$, or $ZA+ZC=2ZT$. But $ZA+ZC=(ZA-AT)+(ZC+CT)=2ZT$, which completes the proof. There is one other way to solve the problem Let $AM$ meet the line $BC$ at $K$ By Menelaus for triangle $ABK$ easy to see that to prove