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As shown in the figure, the area of square ABCD is 16, △ Abe is an equilateral triangle, point E is in square ABCD, and there is a point P on diagonal BD, so that the sum of PC + PE is minimum, then the minimum value is () A. 4B. 23C. 26D. 2
If ∵ square ABCD, ∵ AC ⊥ BD, OA = OC, ∵ C and a are symmetrical about BD, that is, the symmetry point of C about BD is a, and connecting AE with BD to P, then the value of EP + CP is the smallest, ∵ C and a are symmetrical about BD, ∵ CP = AP, ∵ EP + CP = AE, ∵ equilateral triangle Abe, ∵ EP + CP = AE = AB, and the area of ∵ square ABCD is 16
As shown in the figure, the side length of square ABCD is 1, point P is a point on diagonal BD, connecting CP. when BP is equal to BC, make PE ⊥ PC, intersect AB side with E, then be=
Be is equal to 3
As shown in the figure, point E is a point on the diagonal BD of rectangle ABCD, and be= BC.AB=3 . BC = 4. Point P is the point on the line EC, and PQ ⊥ BC is at point Q. PR ⊥ BD is at point R
(1) It is easy to prove that PR + PQ =?
(2) If P is any point on EC (not coincident with E.C.) and other conditions remain unchanged, then the conclusion in (1) still holds. If it holds, please give proof. If not, please give reasons
(3) When p is any point on the extended line of line segment EC and other conditions remain unchanged, what is the quantitative relationship between PR and PQ
1) PR + PQ = AB * BC / BD as EF ⊥ BC intersects BC at point F. connecting BP, ∵ BEP area = 1 / 2be * PR, ∵ BCP area = 1 / 2BC * PQ, be = BC ∵ BCE area = 1 / 2BC * (PR + PQ) ∵ BCE area = 1 / 2BC * EF, ∵ PR + PQ = EF ∵ EF ⊥ BC, CD ⊥ BC
In the cube abcd-a'b'c'd 'with edge length a, e, F, P, q are the midpoint of BC, c'd', ad '. BD respectively. The proof of (1) PQ / / plane dcc'd'
(2) Find the length of PQ
(3) Prove EF / / plane bb'd'd
(1) Because P is the midpoint of ad ', q is the midpoint of BD, and it is also the midpoint of AC. according to the median line theorem of triangle, PQ / / cd'pq is not in plane dcc'd', so PQ / / plane dcc'd '(2) CD' = root 2pq = 1 / 2CD '= root 2 / 2 (3) connects QD', EF, QE because QE / / d'f and QE = d'f, that is, the quadrilateral qed'f is a parallelogram, so EF / / D
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