answer:To prove that ( A_{1}C_{1} = B_{1}D_{1} ), ( A_{1}^{prime}C_{1}^{prime} = B_{1}^{prime}D_{1}^{prime} ), and ( A_{1}^{prime prime}C_{1}^{prime prime} = B_{1}^{prime prime}D_{1}^{prime prime} ), we proceed as follows: 1. **Calculate ( BA_{1} )**: Given that point ( A_{1} ) is the point where the incircle of ( triangle ABD ) touches ( BD ), by the properties of the incircle: [ BA_{1} = frac{1}{2}(AB + BD - AD) ] 2. **Calculate ( C_{1}D )**: Given that point ( C_{1} ) is the point where the incircle of ( triangle BCD ) touches ( CD ), by the properties of the incircle: [ C_{1}D = frac{1}{2}(CD + BD - BC) ] 3. **Determine ( A_{1}C_{1} )**: [ A_{1}C_{1} = |BA_{1} - BD + C_{1}D| ] Substitute the expressions for ( BA_{1} ) and ( C_{1}D ) derived in steps 1 and 2: [ A_{1}C_{1} = left| frac{1}{2}(AB + BD - AD) - BD + frac{1}{2}(CD + BD - BC) right| ] Simplify the expression inside the absolute value: [ A_{1}C_{1} = left| frac{1}{2}(AB + BD - AD) + frac{1}{2}(CD + BD - BC) - BD right| ] [ A_{1}C_{1} = left| frac{1}{2} (AB + BD - AD + CD + BD - BC) - BD right| ] [ A_{1}C_{1} = left| frac{1}{2} (AB + CD - AD - BC + 2BD) - BD right| ] [ A_{1}C_{1} = left| frac{1}{2} (AB + CD - AD - BC) right| ] 4. **Calculate ( B_{1}D_{1} )**: By similar reasoning, using points ( B_{1} ) and ( D_{1} ) with the incircles of ( triangle BCD ) and ( triangle DAB ), respectively: [ B_{1} = frac{1}{2}(BC + CD - BD) ] [ D_{1} = frac{1}{2}(DA + AD - BD) ] We calculate ( B_{1}D_{1} ) analogously: [ B_{1}D_{1} = |BD_{1} - BD + D_{1}| ] [ B_{1}D_{1} = left| frac{1}{2} (BC + AD - BD) + frac{1}{2}(DA + AD - BD) - BD right| ] Simplify as before: [ B_{1}D_{1} = left| frac{1}{2} (AB + CD - AD - BC) right| ] Since both expressions for ( A_{1}C_{1} ) and ( B_{1}D_{1} ) are identical, we have: [ A_{1}C_{1} = B_{1}D_{1} ] Similarly, by the same logic, we can prove: [ A_{1}^{prime}C_{1}^{prime} = B_{1}^{prime}D_{1}^{prime} ] and [ A_{1}^{prime prime}C_{1}^{prime prime} = B_{1}^{prime prime}D_{1}^{prime prime} ] [ boxed{A_{1}C_{1} = B_{1}D_{1}, , A_{1}^{prime}C_{1}^{prime} = B_{1}^{prime}D_{1}^{prime}, , A_{1}^{prime prime}C_{1}^{prime prime} = B_{1}^{prime prime}D_{1}^{prime prime}} ]

answer:We first calculate the product of 102 and 104: [ 102 times 104 = 10608. ] Next, we need to find the remainder when 10608 is divided by 8. We do this by calculating: [ 10608 div 8 = 1326. ] Since 1326 is an integer and the division leaves no remainder, (102 times 104) is also a multiple of 8. Hence, the remainder when it is divided by 8 is [ boxed{0}. ]

answer:From the problem statement, we first rewrite the expression: [ frac{cos frac{P - Q}{2} cos frac{R}{2} - sin frac{P - Q}{2} sin frac{R}{2}}{sin frac{R}{2} cos frac{R}{2}}. ] The numerator simplifies to: [ cos left(frac{P - Q}{2} + frac{R}{2}right) = cos frac{P - Q + R}{2} = cos frac{180^circ - Q - Q}{2} = cos (90^circ - Q) = sin Q. ] The denominator is frac{1}{2} sin R. By the Law of Sines, the expression becomes: [ frac{2 sin Q}{sin R} = frac{2 cdot PR}{PQ} = frac{16}{7}. ] Thus, the final result is: [ boxed{frac{16}{7}}. ]

answer:(1) f(x)=2sin xcos x+2sqrt{3}cos^{2}x-sqrt{3}=sin 2x+sqrt{3}cos 2x=2sin(2x+frac{pi}{3}), Thus, the smallest positive period of f(x) is T=frac{2pi}{2}=pi. From 2kpi+frac{pi}{2}leq 2x+frac{pi}{3}leq 2kpi+frac{3pi}{2}, we get kpi+frac{pi}{12}leq xleq kpi+frac{7pi}{12}, kinmathbb{Z}, So, the interval where f(x) is decreasing is xin[kpi+frac{pi}{12}, kpi+frac{7pi}{12}] (kinmathbb{Z}); (2) From f(frac{A}{2}-frac{pi}{6})=2sin(2(frac{A}{2}-frac{pi}{6})+frac{pi}{3})=2sin A=sqrt{3}, And A is an acute angle, thus A=frac{pi}{3}. By the law of sines, we get 2R=frac{a}{sin A}=frac{7}{frac{sqrt{3}}{2}}=frac{14}{sqrt{3}}, sin B+sin C=frac{b+c}{2R}=frac{13sqrt{3}}{14}, So, b+c=frac{13sqrt{3}}{14}cdotfrac{14}{sqrt{3}}=13, By the cosine law, we know cos A=frac{b^{2}+c^{2}-a^{2}}{2bc}=frac{(b+c)^{2}-2bc-a^{2}}{2bc}=frac{1}{2}, We can find bc=40, Hence, the area of triangle ABC is S_{triangle ABC}=frac{1}{2}bcsin A=boxed{10sqrt{3}}.