answer:Consider the right-angled triangle ( triangle ABC ) with the right angle at ( C ). Points ( D ) and ( E ) are taken on segments ( AC ) and ( CB ) respectively. We need to prove that the perpendiculars from ( C ) to ( DE ), ( EA ), ( AB ), and ( BD ) are concurrent. Refer to the diagram below: [ begin{tikzpicture} coordinate (C) at (0,0); coordinate (A) at (4,0); coordinate (B) at (0,3); coordinate (D) at (2,0); coordinate (E) at (0,2); coordinate (M) at (2,1.5); coordinate (T) at (1,2.5); draw [name path=Ab] (A) -- (B); draw (A) -- (C) -- (B) -- (A); draw (D) -- (E); draw [name path=De] (D) -- (E); draw [name path=Ea] (E) -- (A); draw [name path=Ab] (A) -- (B); draw [name path=Bd] (B) -- (D); path [name intersections={of=Ea and Bd, by=M}]; path [name intersections={of=De and Ab, by=K}]; draw (C) -- (T) -- (K) -- (C) -- (M); path [name intersections={of=Ea and De, by=T}]; draw (C) -- (P); filldraw[black] (C) circle (2pt) node[anchor=east] {C}; filldraw[black] (A) circle (2pt) node[anchor=south east] {A}; filldraw[black] (B) circle (2pt) node[anchor=south east] {B}; filldraw[black] (D) circle (2pt) node[anchor=north east] {D}; filldraw[black] (E) circle (2pt) node[anchor=north east] {E}; filldraw[black] (M) circle (2pt) node[anchor=north east] {M}; filldraw[black] (T) circle (2pt) node[anchor=north east] {T}; end{tikzpicture} ] Let's denote: - ( angle DBC = alpha ) - ( angle EAC = beta ) From geometrical properties, we have: - ( angle DCT = alpha ) - ( angle MCE = beta ) Now, consider the cyclic quadrilateral formed by points ( C, T, K, B ). According to the inscribed angle theorem, we know: [ angle TKC = angle TBC = alpha ] Similarly, for quadrilateral ( PKMC ): [ angle CKM = angle CAM = beta ] Additionally, since ( angle DPT = angle DCT = alpha ) and ( angle EPM = angle ECM = beta ), it follows: [ angle TKM + angle TPM = (alpha + beta) + (180^circ - alpha - beta) = 180^circ ] Thus, by the property that opposite angles of a cyclic quadrilateral sum up to ( 180^circ ), we conclude that points ( P, M, K, T ) are concyclic. Therefore, the perpendiculars from ( C ) to ( DE ), ( EA ), ( AB ), and ( BD ) are concurrent at the orthocenter of ( triangle PMKT ). [ boxed{text{The perpendiculars from ( C ) to ( DE ), ( EA ), ( AB ), and ( BD ) are concurrent.}} ]

answer:Given the total area of the region formed by the eight squares is 648 square centimeters, finding the area of a single square involves dividing the total area by the number of squares: [ frac{648}{8} = 81 text{ square centimeters} ] Assuming all squares are congruent and perfect squares, the side length of one square can be calculated by taking the square root of the area of one square: [ sqrt{81} = 9 text{ centimeters} ] To determine the perimeter, visualize the "plus" shape: one center square with a square extending outward on each side (top, bottom, left, right), amounting to five outward extensions: - There are 4 sides extended outward from the central square, of which 2 are external on each side. - The entire outer boundary of the plus sign has (12) sides, given each open face of the central and outward extension squares contributes to the boundary. Therefore, the perimeter of the whole region would be calculated as: [ 12 times 9 = 108 text{ centimeters} ] So, the perimeter of the region is ( boxed{108 text{ centimeters}} ).

answer:**Answer:** From the graph and properties of the logarithmic function y=log_{0.7}x, we know that log_{0.7}6 < 0. From the graph and properties of the exponential functions y=0.7^x and y=6^x, we know that 0.7^6 < 1 and 6^{0.7} > 1. Therefore, log_{0.7}6 < 0.7^6 < 6^{0.7}. Hence, the correct option is boxed{text{D}}. **Analysis:** By understanding the graph and properties of the logarithmic function, we can conclude that log_{0.7}6 < 0. Then, by analyzing the properties of the exponential functions, we find that 0.7^6 < 1 and 6^{0.7} > 1, leading to the conclusion.

answer:The vector bold{v} can be expanded as: [bold{v} = begin{pmatrix} 3 + 8t -2 + 4t 1 - 2t -4 + 6t end{pmatrix}.] We need to find out when bold{v} is closest to bold{a}. This happens when the vector from bold{v} to bold{a} is orthogonal to the direction vector begin{pmatrix} 8 4 -2 6 end{pmatrix}. Set up the orthogonality condition: [left( begin{pmatrix} 3 + 8t -2 + 4t 1 - 2t -4 + 6t end{pmatrix} - begin{pmatrix} 5 3 7 2 end{pmatrix} right) cdot begin{pmatrix} 8 4 -2 6 end{pmatrix} = 0.] Perform the subtraction and dot product: [begin{pmatrix} -2 + 8t -5 + 4t -6 - 2t -6 + 6t end{pmatrix} cdot begin{pmatrix} 8 4 -2 6 end{pmatrix} = 0,] [ (-2 + 8t)(8) + (-5 + 4t)(4) + (-6 - 2t)(-2) + (-6 + 6t)(6) = 0. ] Simplifying: [ -16 + 64t - 20 + 16t + 12 + 4t + (-36 + 36t) = 0, ] [ 120t - 60 = 0, ] [ t = frac{60}{120} = boxed{frac{1}{2}}. ]