As shown in the figure, ab ‖ CD, OA = OD, points F, D, O, a, e are on the same line, AE = DF, verification: EB ‖ CF

As shown in the figure, ab ‖ CD, OA = OD, points F, D, O, a, e are on the same line, AE = DF, verification: EB ‖ CF

Proof: ∵ ab ∥ CD,
∴∠DCO=∠ABO，∠CDO=∠BAO，
In △ AOB and △ doc,
∠ABO＝∠DCO
∠BAO＝∠CDO
OA＝OD ，
∴△AOB≌△DOC（AAS），
∴OC=OB，
∵OA=OD，AE=DF，
/ / OA + AE = od + DF, that is, OA = of,
In △ COF and △ BOE,
OC＝OB
∠COF＝∠BOE
OF＝OE ，
∴△COF≌△BOE（SAS），
∴∠F=∠E，
∴BE∥CF．

As shown in the figure, the extension lines of the chords AB and CD of the center O intersect with the point P. if the arc AC = 130 degrees and the arc DB = 30 degrees, what degrees is the angle P equal to

Connect BC, then ∠ ABC = 130 ° / 2 = 65 °
∠PCB=30°/2=15°
If the outer angle is equal to the sum of two interior angles which are not adjacent to it, then
∠P=65°-15°=50°

It is known that: as shown in the figure, BD is the diagonal of the parallelogram ABCD, O is the midpoint of BD, EF ⊥ BD is at point O, and it intersects with AD and BC at points E and f respectively. Verification: de = DF

It is proved that in the parallelogram ABCD, ad ∥ BC,
∴∠OBF=∠ODE
∵ o is the midpoint of BD
∴OB=OD
In △ BOF and △ doe,
A kind of
∠OBF＝∠ODE
OB＝OD
∠BOF＝∠DOE
∴△BOF≌△DOE
∴OF=OE
∵ EF ⊥ BD at point o
∴DE=DF．

As shown in the figure, O is the midpoint of the diagonal AC of the parallelogram ABCD, EF passes through point O, intersects ad at point E, crosses BC at point F, connects be and DF, tries to explain that the quadrilateral BEDF is a parallelogram

∵▱ABCD，
∴AD∥
.
.
CB，OA=OC．
∴∠EAO=∠FCO．
And ∵ AOE = ∵ COF,
∴△AOE≌△COF．
∴AE=CF．
∵AD∥
.
.
BC，
∴（AD-AE）∥
.
.
(bc-cf), i.e. de ‖
.
.
BF．
The quadrilateral BEDF is a parallelogram

As shown in the figure, given that AD and BC intersect at point O, OA = OD, angle a = angle D, try to judge whether AB and CD are equal

prove:
∵ AOB and  COD are opposite vertex angles
∴∠AOB＝∠COD
∵∠A＝∠D,OA＝OD
∴△AOB≌△DOC （ASA）
∴AB＝CD

Take two points a and B on the straight line m so that ab = 10cm, and then take a point P on m so that PA = 2cm, m and N are the midpoint of PA and Pb respectively. Find the length of line Mn

As shown in the figure, (1) when point P is on the line segment AB, Pb = ab-pa = 8cm, m and N are the midpoint of PA and Pb respectively, ﹥ Mn = PM + PN = 12ap + 12bp = 1 + 4 = 5 (CM); (2) when point P is on the extension line of line Ba, Pb = AB + PA = 12cm, m and N are the midpoint of PA and Pb respectively,  Mn = pn-pm = 12bp-12ap = 6-1 = 5

It is known that point AB is two points on the line AB, and ab = 10, point P is a point on ray BA (point P does not coincide with ab), M is the midpoint of PA, n is the midpoint of Pb, and find the line segment Mn

Line Mn = 1 / 2Ab = 5;
In case 1, when point P is between AB, Mn = MP + PN = 1 / 2AP + 1 / 2PB = 1 / 2 (AP + Pb) = 1 / 2Ab = 5;
In case 2, when the point P is outside AB, we can also obtain Mn = 5

(see supplement) as shown in the figure, point P is the point outside the line Mn, PD ⊥ Mn, the perpendicular foot is D, a and B are two points on the line Mn, connecting PA and Pb, and PA = 4cm As shown in the figure, point P is a point outside the line Mn, PD ⊥ Mn, the perpendicular foot is D, a and B are the two points on the line Mn, connecting PA and Pb, known as PA = 4cm, Pb = 5cm, PD = 3cm, then the distance from point P to line Mn is () a, 4cm B, 5cm C, 3cm D, which cannot be determined

PD is not vertical to Mn! Then the distance between P and Mn is not 3cm, select C

It is known that a and B are on both sides of the straight line L. find a point on L so that PA + Pb is the smallest. (as shown in the figure)

The intersection point between the two points and the straight line is the point P,
So PA + Pb is the smallest,
The reason is that the line is the shortest between two points

As shown in the figure, two triangular plates with the same shape and size, including 30 degrees and 60 degrees, are placed as shown in the figure, PA.PB It coincides with the straight line Mn, and the triangle plate PAC, The triangular plate PBD can rotate anticlockwise around point P

(1) Angle DPC = 180 ° - 30 ° - 60 ° = 90 °
(2) Angle DPB = 30 degrees
Angle APC = 60 degrees
Angle EPF = angle EPD - angle EPD
=Angle APD divided by 2 - angle cpd divided by 2
=30°