(1 / 2) in the pyramid p-abcd, the bottom surface ABCD is rectangular, the vertical plane ABCD of PA, PA = ad = 1, ab = 2, e and F are the midpoint of AB and PD respectively（

Take the midpoint m of PC and connect FM and em
FM / / DC and FM = 1 / 2dc
And because of the rectangle ABCD
So DC is parallel and equal to ab
So FM / / AB and FM = 1 / 2Ab
And because e is the midpoint of ab
So AE is parallel and equal to FM
So parallelogram fmae
So fa / / me
And because me ∈ face PEC
So AF / PEC

Point m (1,0) and n (- 1,0), point P is a moving point on the line 2x-y-1 = 0, then what is the minimum value of the square of PM plus the square of PN

Let p be (m, n), PM's Square plus PN's Square is the smallest, then M and N satisfy the equation 2x-y-1 = 0, so n = 2m-1, PM's Square plus PN's Square = (m-1) × (m-1) + (n-0) × (n-0) + (M + 1) × (M + 1) + (n-0) × (n-0) = m × m-2m + 1 + m × m + 2m + 1 + 2n × n =

Given m (1,0) and n (- 1,0), point P is a moving point on the line 2x-y-1 = 0, then the minimum value of | PM | ^ 2 + | PN | ^ 2 is
I don't understand the problem at all

Let P (x, 2x-1)
Then | PM | ^ 2 + | PN | ^ 2
=(x-1)²+(2x-1)²+(x+1)²+(2x-1)²
=10x²-8x+4
It's a quadratic function of X with the opening up
Axis of symmetry x = 2 / 5
When x = 2 / 5, there is a minimum value of 12 / 5

AB is the diameter of circle O, and ab = 10. If the length of chord Mn is 8, the two ends of Mn slide on the circumference and always intersect ab. the distance from counting points a and B to Mn is H1, H2, and / h1-h2 is calculated

As shown in the figure, the diameter ab of the circle O and the chord Mn intersect at the point P, ab = 10, Mn = 8. The distances from points a and B to Mn are AC = H1, BD = H2 respectively. Connect OM, on, through O, make OE perpendicular to Mn, perpendicular to e, then E is the midpoint of Mn. In the right triangle OEM, OM = 5, EM = 4, so OE = 3

In △ ABC, ab = AC, P and Q are points on AC and ab respectively, AP = PQ = BP = CB
Find the value of ∠ a (∠ A is the top angle)
It's about 45 degrees, but I don't know how to prove it
The faster the better

AP = PQ = BP Q seems to be contradictory in ab

In △ ABC, be ⊥ AC, ad ⊥ BC, P is on AC, and AP = ad, through P as PQ, parallel CB intersects AB with Q, verification: PQ = be

No picture

Equilateral triangle ABC, P is a point on AB, q is a point on AC, AP = CQ, PC, PQ connected, the midpoint m on PQ connected am, am = 19 find CP =?

In positive △ ABC, P is a point on AB, q is a point on AC, and AP = CQ
The distance from point P to point C is equal to?

Is it a question to fill in the blanks? Yes, from the most special case, that is, PQ is ab respectively, and the midpoint of AC is PC = 38cm

Equilateral triangle ABC, P is a point on AB, q is a point on AC, AP = CQ, connect the midpoint m on PQ, connect am, am = 19, find CP

As shown in the figure, ab = a, BC = B, CD = C, de = D, EF = e. the sum of the lengths of all line segments with a, B, C, D, e and F as the endpoints is______ ．

AB = a, AC = a + B, ar = a + B + C, A5 = a + B + C + R, AF = a + B + C + R + 5, BC = B, Br = B + C, B5 = B + C + R, BF = B + C + R + 5, Cr = C, C5 = C + R, CF = C + R + 5, R5 = R, RF = R + 5

(1 / 2) in the pyramid p-abcd, the bottom surface ABCD is rectangular, the vertical plane ABCD of PA, PA = ad = 1, ab = 2, e and F are the midpoint of AB and PD respectively（