In the triangle ABC, the point O is the previous moving point of AC, and the crossing point O is a s

1.Angular OEC = Angular BCE = Angular Eco, OE = OC, Angular OFC = Angular OCF, OF=OC, EO = FO.

When AO=CO, the quadrilateral AECF is a parallelogram, the angle ECF=90 degrees, the quadrilateral AECF is a rectangle, and when O moves to the midpoint of AC, the quadrilateral AECF is a rectangle.

3.In 2, the angle AEC=90 degrees, if the quadrilateral AECF is a square, then AE=EC, that is, the angle ACE=angle CAE=45 degrees, then there is an angle ACE+angle BCE=90 degrees, so when the angle C in the triangle ABC is a right angle, the quadrilateral AECF is a square.

Parallel to BC, the internal misalignment angles are equal, and the property of bisector of two angles.

Knowable. Angular ECO = Angular CEF EO = OC

The same can be said: fo=oc

So eo=fo

2 Rectangles have one property: the diagonals are equal and bisected.

It can also be shown that a quadrilateral with equal and bisected diagonals is a rectangle.

So when oa=oc, i.e., o is the midpoint of ac, the quadrilateral aecf is rectangular.

3 If the quadrilateral AECF is required to be a square, then based on (2), only AC is required to be perpendicular to MN

That is, AC is perpendicular to BC, so the angle C of ABC is a right angle!

As follows:

1. Proof. mn//bc

oec=∠bce

ofc=∠fcg

BCE= OCE (OE is the bisector of the internal angle of BCA) OEC= OCE

OE = OC OCF = FCG (OF is the outer angle bisector of BCA) OCF = OFC

of=ocoe=of。

2. When the O point is at the AC midpoint, the quadrilateral AECF is rectangular.

Oe=of and oc=oa from 1 (O is the midpoint of AC), so the quadrilateral AECF is a parallelogram (a quadrilateral with diagonals bisected by each other is a parallelogram).

And because the angle BCA + angle ACK = 180 degrees (K is a point on the BC extension line), the angle BCE = angle ECA and the angle BCE + angle ECA = angle BCA, the angle ACF = angle FCK and the angle ACF + angle FCK = angle ACK, so the angle ECF = angle ECA + angle ACF = 1 2BCK = 90 degrees.

So the quadrilateral acef is rectangular (there is a parallelogram with a right angle that is a rectangle).

The common methods for determining a rectangle are as follows:

1) There is a parallelogram with an angle of right angles and a rectangle.

2) A parallelogram with equal diagonals is a rectangle.

3) There are three corners that are right angles, and the quadrilateral is rectangular.

4) Theorem: It has been proved that in the same plane, any two angles are right angles, and any set of quadrilaterals with equal opposite sides is a rectangle.

5) Quadrilaterals with equal diagonals and bisected from each other are rectangulars.

1. Solution: Because mn bc

So angular bce = angular fec

And because the angle BCE = angle ECO (CE is the angle bisector of the angle BCA), in the triangle OEC, the angle OEC = angle OCE then OCE is an isosceles triangle, that is, OE = OC

The same can be said for of=oc

then there is oe=oc=of

i.e. oe=of

2. Solution: When the O point is at the midpoint of AC, the quadrilateral AECF is a rectangle and obtains OE=OF from 1

and oc=oa (o is the midpoint of ac).

So the quadrilateral AECF is a parallelogram (a quadrilateral whose diagonals are bisected by each other is a parallelogram).

And because the angle BCA + angle ACK = 180 degrees (K is a point on the BC extension line), the angle BCE = angle ECA and the angle BCE + angle ECA = angle BCA angle ACF = angle FCK and the angle ACF + angle FCK = angle ACK, so the angle ECF = angle ECA + angle ACF = 1 2BCK = 90 degrees, so the quadrilateral ACEF is rectangular (there is a parallelogram with a right angle that is a rectangle).

Proof:

MN BC,CE BIX, ACB, CF BIX, ACD BCE= ACE= OEC, OCF= FCD= ofcoe=oc,OC=of

oe=ofacb+∠acd=2∠ace+2∠acf=180°∠ace+∠acf=∠ecf=90°

To prove that a quadrilateral AECF is a rectangle, it is only necessary to prove that it is a parallelogram.

OE=OFWhat OA=OC, the quadrilateral AECF is a parallelogram and is also a rectangle.

Judgment method: 1. Acute triangle: the maximum angle of the three inner angles of the triangle is less than 90 degrees.

2. Right triangle: The maximum angle of the three inner angles of the triangle is equal to 90 degrees.

3. Obtuse triangle: The maximum angle of the three inner angles of the triangle is greater than 90 degrees and less than 180 degrees.

Among them, acute triangles and obtuse triangles are collectively called oblique triangles.

Your diagram is just too ugly.

1) Take any point of D on the BC extension line

CE bisects ACB, OCE = ECB

Mn BC, ecb= oec= oceoe=oc, and the same can be proved: of=oc

oe=of2)∠acb+∠acd=2∠ace+2∠acf=180°∠ace+∠acf=∠ecf=90°

To prove that a quadrilateral AECF is rectangular. It is only necessary to prove that it is a parallelogram.

OE=OFWhat OA=OC, the quadrilateral AECF is a parallelogram and is also a rectangle.

As shown in the figure, in ABC, the point O is a moving point on the edge of AC, and the crossing point O is a straight line Mn BC, and the angle bisector of MN intersection BCA is at point E, and the outer angle ACG bisector of intersection BCA is at point F

1) Try to explain eo=fo;

2) When the point o moves to where is the quadrilateral AECF rectangular? and give reasons

3) When the point O moves and what conditions does abc meet, is the quadrilateral AECF a square? and give reasons

Solution: (1) Mn bc, OEC= BCE, ofc= gcf, and CE bisects bco, cf bisects gco, oce= bce, ocf= gcf, oce= oec, ocf= ofc, eo=co, fo=co, eo=fo

2) When the point O moves to the midpoint of AC, the quadrilateral AECF is rectangular

When the point O moves to the midpoint of AC, ao=co, and eo=fo, the quadrilateral AECF is a parallelogram, FO=CO, AO=CO=EO=FO, AO+CO=EO+FO, i.e., AC=EF, and the quadrilateral AECF is a rectangle

3) When the point O moves to the midpoint of AC, and ABC satisfies ACB as a right-angled triangle, the quadrilateral AECF is a square

From (2), it is known that when the point O moves to the midpoint of AC, the quadrilateral AECF is a rectangle, and MN BC is known, when ACB=90°, then AOF= CoE= COF= AOE=90°, AC EF, and the quadrilateral AECF is a square

CE and CF are angular bisectors.

Angular OCF = Angular DCF

Angular oce = angular ecb

So the angle ecf = 90 degrees.

mn//bc

So the angle dcf = angle ofc = ocf

Angular OCE = Angular OEC = Angular ECB

So the side oe=oc=of(isosceles 3 angular rows).

Because no matter how the point O moves, of=oc=oe is true, and the angle ecf=90 degrees.

Counter-argument, when the AECF is rectangular.

So ac=ef (diagonal equals in the rectangle).

ac=ao+oc

ef=eo+of

of=oc=oe

So we get of=oc=oe=ao

So when o is the ac midpoint, it's rectangular.

CE and CF are.

Angular OCF = Angular DCF

Angular oce = angular ecb

So the angle ecf = 90 degrees.

mn//bc

So the angle dcf = angle ofc = ocf

Angular OCE = Angular OEC = Angular ECB

So the side oe=oc=of(isosceles 3 angular rows).

Because no matter how the point O moves, of=oc=oe is true, and the angle ecf=90 degrees.

Counter-argument, when the AECF is rectangular.

So ac=ef (diagonal equals in the rectangle).

ac=ao+oc

ef=eo+of

of=oc=oe

So we get of=oc=oe=ao

So when o is the ac midpoint, it's rectangular.

1. Solution: Because mn bc

So angular bce = angular fec

And because the angle BCE = angle ECO (CE is the angle bisector of the angle BCA), in the triangle OEC, the angle OEC = angle OCE then OCE is an isosceles triangle, that is, OE = OC

The same can be said for of=oc

then there is oe=oc=of

i.e. oe=of

2. Solution: When the O point is at the midpoint of AC, the quadrilateral AECF is a rectangle and obtains OE=OF from 1

and oc=oa (o is the midpoint of ac).

So the quadrilateral AECF is a parallelogram (a quadrilateral whose diagonals are bisected by each other is a parallelogram).

And because the angle BCA + angle ACK = 180 degrees (K is a point on the BC extension line), the angle BCE = angle ECA and the angle BCE + angle ECA = angle BCA angle ACF = angle fck and angle ACF + angle FCK = angle ACK, so the angle ECF = angle ECA + angle ACF = 1 2BCK = 90 degrees, so the quadrilateral ACEF is a rectangle (there is a parallelogram with a right angle is a rectangle), if you are satisfied, remember to adopt it!

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(1) EF BC, OEC= ECB, and CE is the ACB bisector, ECB= OCE

oce=∠oec

Launch of OE=OC

In the same way, oc=of (there is another way to prove that the sum of the inner and outer angles is 180°, and the sum is 90° after bisecting, ECF is a right angle, and oe=oc, theorem: the middle line of the hypotenuse of a right triangle is equal to half of the hypotenuse, and its inverse theorem can be used to obtain that oc is the middle line).

OE=OF2) When O moves to the midpoint of AC, the quadrilateral AECF is rectangular.

Analysis process: ecf = 90°, just prove that the quadrilateral aecf is a parallelogram and conclude that it is a rectangle (theorem: a parallelogram with an angle is a right angle is a rectangle. ）

Proof process: O is the midpoint of EF (proven above), O is the midpoint of AC, and the quadrilateral AECF is a parallelogram (theorem: a quadrilateral with diagonals bisected by each other is a parallelogram).

and ecf=90°, the quadrilateral aecf is rectangular.

Proof: EF BC

e=∠bce

f= fck (k is a point on the BC extension line, which is used to determine this angle) and CE bisects acb cf bisects ack

bce=∠ace

acf=∠fck

e=∠ace

f=∠acf

oe =oc

oc=of∴oe=of

2. When O moves to the midpoint of AC, AECF is rectangular.

Proof: OE=OF OA=OC EF AC is the diagonal of AECF.

ACEF is a parallelogram.

EC and CF are bisectors of ACB and ACK.

ec cf, then one of the internal angles of a parallelogram is 90°

Then this quadrilateral is rectangular.

Because EF BC, then the angle OEC = angle ECB, CE is the angle ACB bisector, then the angle ECB = angle OCE

So the angle oce = angle oec, then oe = oc, and the same way can get oc = of

i.e. oe=of

1. All proofs:1,

EC bisects BCA

eca＝∠ecb＝∠bca/2

FC bisects ACG

fca＝∠fcg＝∠acg/2

eca+∠fca＝∠bca/2+∠acg/2＝（∠bca+∠acg）/2

bca+∠acg＝180

eca+∠fca＝180/2＝90

ecf＝∠eca+∠fca＝90

mn∥bc∠ofc＝∠fcg

ofc＝∠fca

of＝ocmn∥bc

oec＝∠ecb

oec＝∠eca

oe＝ocoe＝of

2. When O is at the midpoint of AC, AECF is rectangular.

o is the midpoint of AC.

ao＝cooe＝of，∠aoe＝∠cof

AOE is equal to COF

The same can be said for AE CF and AF CE

Parallelogram AECF

ecf＝90

Rectangular AECF