answer:1. **Problem Setup:** - Given triangle ABC with AC neq BC. - Point D is inside triangle ABC such that angle ADB = 90^circ + frac{1}{2} angle ACB. - The circumcircle of triangle ABC intersects line AB at point P such that P neq A, B. - The circumcircle of triangle ADC intersects line AD at point Q such that Q neq A, D. - We need to prove that line PQ bisects angle BPC. 2. **Internal Angle Bisectors and Ex-circle Connections:** - Let I be the incenter of triangle ABC. - The internal bisector of angle angle ACB intersects AB at L and the circumcircle triangle ABC again at point M. - By properties of the internal bisector, angle AIB = 90^circ + frac{1}{2} angle ACB = angle ADB. - Thus, points A, I, D, B are concyclic, meaning they lie on the same circle Gamma. 3. **Circumcircle Intersections:** - M is the center of circle Gamma that passes through A, I, D, B. - By the power of point from circle properties, PC^2 = PA cdot PB and QC^2 = QA cdot QD. - Both P and Q are points of tangency to circles involving C and corresponding lines. 4. **Nature of Point P and Q:** - Points P, Q lie along the radical axis of circles triangle ABC and triangle ADC. - Since these points lie along the radical axis, PQ perp CL where L is defined above. 5. **Angle Calculation:** - At point C: [ angle PCL = angle PCA + frac{1}{2} angle ACB ] - For angles inside triangle ABC: [ angle PCA = angle CBA, quad angle ACB text{ as given and related previously} ] - Therefore: [ angle PCL = angle CBA + frac{1}{2} angle ACB = angle CLP ] - This implies PL = PC. 6. **Conclusion:** - Hence, line PQ bisects angle BPC as required. [ boxed{} ]

answer:To determine the sum of the first 100 terms of the sequence, we need to understand its pattern. Let's start by verifying the terms of the sequence manually according to the recurrence relation (a_{n} = a_{n-1} - a_{n-2}): 1. The first term is given as: [ a_1 = 1 ] 2. The second term is given as: [ a_2 = 3 ] 3. The third term is: [ a_3 = a_2 - a_1 = 3 - 1 = 2 ] 4. The fourth term is: [ a_4 = a_3 - a_2 = 2 - 3 = -1 ] 5. The fifth term is: [ a_5 = a_4 - a_3 = -1 - 2 = -3 ] 6. The sixth term is: [ a_6 = a_5 - a_4 = -3 - (-1) = -3 + 1 = -2 ] 7. The seventh term is: [ a_7 = a_6 - a_5 = -2 - (-3) = -2 + 3 = 1 ] 8. The eighth term is: [ a_8 = a_7 - a_6 = 1 - (-2) = 1 + 2 = 3 ] 9. The ninth term is: [ a_9 = a_8 - a_7 = 3 - 1 = 2 ] Observing the terms, we see that (a_7 = a_1), (a_8 = a_2), and (a_9 = a_3). The sequence starts to repeat every 6 terms. We can summarize the sequence as: [ 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, ldots ] To find the sum of the first 100 terms: 1. First, verify the sum of one complete cycle of 6 terms: [ 1 + 3 + 2 - 1 - 3 - 2 = 0 ] 2. Since the sequence repeats every 6 terms and 100 terms is (leftlfloor frac{100}{6} rightrfloor = 16) full cycles, plus 4 extra terms. The sum of 96 terms (16 cycles of 6 terms each) is: [ 16 times 0 = 0 ] 3. Now, calculate the sum of the first 4 terms of the next cycle: [ 1 + 3 + 2 - 1 = 5 ] Thus, the sum of the first 100 terms of the sequence is: (boxed{5}).

answer:(I) The function f(x)=2 sqrt {3}cos ^{2}ωx+sin 2ωx- sqrt {3} can be rewritten as f(x) = 2sin (2ωx+ frac {π}{3}) (where ω > 0). Its smallest positive period is given by frac {2π}{2ω}=2π, thus ω= frac {1}{2}. Therefore, f(x)=2sin (x+ frac {π}{3}). (II) By compressing the abscissa of each point on the graph of the function y=f(x) to frac {1}{2} of the original, and keeping the ordinate unchanged, we obtain the graph of the function y=g(x)=2sin (2x+ frac {π}{3}). Let 2kπ- frac {π}{2}leqslant x+ frac {π}{3}leqslant 2kπ+ frac {π}{2} (where kin mathbb{Z}), we find kπ- frac {5π}{12}leqslant xleqslant kπ+ frac {π}{12}. Thus, the monotonically increasing interval of the function g(x) is boxed{[kπ- frac {5π}{12},kπ+ frac {π}{12}]} (where kin mathbb{Z}).

answer:Gwen received 14 dollars and spent 8 dollars. To find out how much money she has left, we subtract the amount she spent from the amount she received: 14 dollars - 8 dollars = 6 dollars Gwen has boxed{6} dollars left.