In the isosceles triangle ABC, AC = BC = a, point P is a point outside the triangle ABC, PA = Pb = PC = root 2A In the isosceles triangle ABC, AC = BC = a, point P is a point outside the triangle ABC, PA = Pb = PC = (radical 2) a, is it wrong to prove the vertical plane ABC of plane PAB? Another question is in the triangular prism abc-a1b1c1, the side edge is perpendicular to the bottom, AC = 3 BC = 4 AB = 5 Aa1 = 4, and point D is the midpoint 1 of ab

In the isosceles triangle ABC, AC = BC = a, point P is a point outside the triangle ABC, PA = Pb = PC = root 2A In the isosceles triangle ABC, AC = BC = a, point P is a point outside the triangle ABC, PA = Pb = PC = (radical 2) a, is it wrong to prove the vertical plane ABC of plane PAB? Another question is in the triangular prism abc-a1b1c1, the side edge is perpendicular to the bottom, AC = 3 BC = 4 AB = 5 Aa1 = 4, and point D is the midpoint 1 of ab

In the first question of trigonometric vertebrae, the normal vectors of two faces are calculated respectively to prove that the dihedral angle of the plane angle of two faces is 90 degrees. In the second question, the known angle BAC is a right angle, AB is the X axis, AC is the axis, Aa1 is the Z axis. The two straight lines are represented by vectors, and the number of the two vectors is multiplied by zero

In the isosceles triangle ABC, there is a point P, PA = 2, Pb = 2, root 3, PC = 4 to find the area of △ ABC

Rotate △ APC clockwise 60 ° around point a to △ AMB, then am = AP = 2, BM = PC = 4, ∠ PAM = 60 °
If PM is connected, then △ PAM is an equilateral triangle, PM = 2
In △ PBM, PM & # 178; + Pb & # 178; = 2 & # 178; + (double root 2) &# 178; = 16
BM²=4²=16
∴PM²+PB²=BM²
Ψ△ PBM is a right triangle, ∠ BPM = 90 °
∴∠APB=90°+60°=150°
If a is used as the extension line of ad ⊥ BP and D is used as the extension line of BP, then APD = 30 degree
∴AD=1,PD=√3
∴AB²=1²+（3√3）²=28
∴AB=2√7
Triangle ABC area = 14 pieces 3 / 2
An equilateral triangle is not an isosceles triangle

If △ PAB is an acute triangle and PC ⊥ AB proves, PA ^ 2 + BC ^ 2 = Pb ^ 2 + AC ^ 2
It is helpful for the responder to give an accurate answer

Pythagorean theorem: PA ^ 2 = ad ^ 2 + PD ^ 2 AC ^ 2 = CD ^ 2 + ad ^ 2 subtraction: PA ^ 2-ac ^ 2 = PD ^ 2-CD ^ 2 similarly: Pb ^ 2-bc ^ 2 = PD ^ 2-CD ^ 2 PA ^ 2-ac ^ 2 = Pb ^ 2-bc ^ 2 PA ^ 2 + BC ^ 2 = Pb ^ 2 + AC ^ 2

As shown in the figure, in the circle O, the chord AB = CD, and the intersection of AB and CD is proved by E.; de = AE

Connect BC, because AB = CD, the arc corresponding to ab = the arc CD corresponding to CD, and the arc ad is the common arc
Arc AB arc ad = arc CD arc ad
Namely: arc BD = arc AC
So: the chord BD corresponding to arc BD = the chord AC corresponding to arc AC
Namely: BD = AC
And because: ab = CD, BC = BC
So triangle ABC congruent triangle DCB
So: ∠ BAC = ∠ CDB
The circle angles corresponding to the equal arc AD are equal, that is: ∠ ACD = ∠ abd
And because: AC = BD
So: ACE congruent triangle DBE
So: AE = de

As shown in the figure, in ⊙ o, the diameter CD intersects the chord AB at point E. if be = 3, AE = 4, de = 2, then the radius of ⊙ o is ()
A. 3B. 4C. 6D. 8

According to the intersecting chord theorem, the radius of a circle is equal to 4 if AE · be = CE · De, and ∵ be = 3, AE = 4, de = 2, ∵ CE = 6 ∵ CD = CE + de = 8

The line L passes through the point a (- 1,1), which is cut by two parallel lines L1: x + 2y-1 = 0 and L2: x + 2y-3 = 0. The midpoint of the line is just on the line L3: x-y-1 = 0. The equation of line L is obtained

The linear equation with equal distance to the parallel lines L1: x + 2y-1 = 0 and L2: x + 2y-3 = 0 is x + 2y-2 = 0. By solving x + 2Y − 2 = 0x − y − 1 = 0, we can get x = 43y = 13, that is, the midpoint of the line segment cut by the parallel lines L1 and L2 is B (43, 13). Therefore, the slope of the line L is k = 13 − 143 + 1 = - 27, and the equation of the line L is Y-1 = - 27 (x + 1)

The coordinates of the midpoint of the chord cut by the circle x2 + y2-2x-2y-6 = 0 are ()
A. （1，0）B. （14，34）C. （34，14）D. （12，12）

If x + y − 1 = 0x2 + Y2 − 2x − 2Y − 6 = 0, we can get 2x2-2x-7 = 0. Let the intersection of a line and a circle be a (x1, Y1), B (X2, Y2), ∧ X1 + x2 = 1, Y1 + y2 = (1-x1) + (1-x2) = 1, ∧ line x + Y-1 = 0 is cut by circle x2 + y2-2x-2y-6 = 0, and the coordinates of the middle point of the chord are (12, 12)

The midpoint of the line segment cut by the circle x2 + y2-2x-2y-7 = 0 is the process point

(1/2,1/2)
Thinking:
x2+y2-2x-2y-7=0
=>(x-1)2+(y-1)2=9
So the circle takes (1,1) as its point and 3 as its radius
Let x + Y-1 = 0 be e, F
The intersection of the vertical line passing through the circle and EF is the midpoint of EF
Let the vertical line passing through the dot be y = x + B, and substitute (1,1) to get b = 0
So y = X
x+y-1=0
The simultaneous equations are x = 1 / 2, y = 1 / 2
That is, the midpoint of the line segment is (1 / 2,1 / 2)

If P (2,1) is known, make a straight line through P, so that the line segment between X + 2y-3 = 0, 2x + 5y-10 = 0 is bisected by P, and solve the linear equation

Let the obtained line segment be AB, and the points a (x1, Y1) and B (X2, Y2) are respectively on the straight line x + 2y-3 = 0, 2x + 5y-10 = 0, ∵ the line segment is bisected by the point P (2, 1), ∵ according to the midpoint formula, X1 + X22 = 2y1 + Y22 = 1; ∵ x2 = 4-x1, y2 = 2-y1, ∵ B (4-x1, 2-y1), the two points are substituted respectively,