Answers to 17 and 18 questions in junior high school math and geometry questions

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

3 O When moving to the midpoint of the AC edge, the quadrilateral AECF is rectangular.

Proof: oe=oc

OE=OF: OA=OC when O is the midpoint of AC

oe=oc=of=oa

The quadrilateral AECF is rectangular.

Reference: junruqu].

Hope by.

17 Solution: (1) Prove: MN crosses the bisector of ACB at the point E, and the bisector of the outer angle of the intersection ACB is at the point F, <>

2=∠5，4=∠6，mn∥bc，∠1=∠5，3=∠6，∠1=∠2，∠3=∠4，eo=co，fo=co，oe=of；

2) Solution: 2 = 5, 4 = 6, 2+ 4 = 5+ 6 = 90°, ce=12, cf=5, ef= 12 5 13

oc=½ef=

3) A: When the point O moves on the edge AC to the midpoint of AC, the quadrilateral AECF is rectangular

Proof that when O is the midpoint of AC, AO=CO, EO=FO, the quadrilateral AECF is a parallelogram, ECF=90°, and the parallelogram AECF is rectangular

18 solutions: 1) Prove: in square ABCD, ab=ad, bae= d=90°, daf+ baf=90°, af be, abe+ baf=90°, abe= daf, in abe and daf, abe daf

ab＝adbae＝∠d

abe≌△daf（asa），af=be；

2) Solution: MP is equal to NQ

**Satisfied***

17 questions. Solution: (1) Mn BC and CE bisected ACB BCE= ACE= CEF (the wrong angles in the parallel lines are equal) OEC is an isosceles triangle.

oc=oe The same goes for oc=of

So oe=of

2）∵∠bcd=180°

ecf = 90° (you understand?) ）

and ce=12, cf=5

EF = 13 (Pythagorean theorem).

oc=1/2ef=

The gods replied so quickly, I will save it.

Solution: Make the auxiliary line EF so that EF AD intersects AC at F because: EF AD

The angle def=60 degrees.

And because ab=ac, angular bac=60

So the triangle ABC is a congruent triangle.

So the outer angle of the angle BCA is 120 degrees.

And because CE is the outer angle bisector of the angle BCA, the angle ACE = 60 degrees.

And because the angle ace = angle CFE + angle fec = angle def = angle dec + angle cef so angle dec = chickium efc

And because the angle cde + angle ced = 60 = angle cfe + angle fec, the angle fec = angle dec = 30 degrees.

Put at an angle cde = 30 degrees.

So cd=ce

So de vertical ac

So the square root of ad=3 *cd

de=2*3/2 square root*cd = square root of 3*cd, so ad=de

Students, please translate into the language of mathematics yourself.

Children's shoes,** not very clear...

Let the side length of regular triangle 2 be x

The sides of the triangle are both x long

The triangle has a side length of x+a

The side length of the triangle is x+2a

The side length of triangle 9 is x+3a

The side length of triangle 9 is the sum of triangles, then x+3a=x+x gives x=3a, and the circumference of the hexagon is 3a+3a+(3a+a)+(3a+a)+(3a+2a)+(3a+2a)+(3a+3a)=30a

1. The square diagonal is also the angular bisector of the angle.

Therefore, the triangle ABP is all equal to the triangle CBP (corner edges), i.e., pc=pa=pe

2. Because, the triangle ABP is equal to the triangle CBP

So, angular PAB = angular PCB, so, angular pad = angular PCD

Because, Pa=PE

So, the angle pae(pad) = the angle e

So, the angle pcd = the angle e

Because, the angle PFC = EFD (equal to the apex angle).

So, angular cpe = angular cde = 90°

3. When the angle abc = 120 °, the angle PDC = 120 °, yes, the angle DPE + the angle DEP = 60 °

Because, the triangle PFC is similar to the triangle DFE

So, angular def = angular pcf. And angular PCF + angular PCB = 60° so, angular dpe = angular PCB

So, angular DPE + angular BPC = angular PCB + angular BPC = 120°, i.e., angular CPE = 60°

Because, pc=pa=pe, the triangle pce is an isosceles triangle because, the angle cpe=60°

So the triangle PCE is an equilateral triangle (an isosceles triangle with an angle equal to 60° is an equilateral triangle).

i.e., ap=pe=pc=ce

Method 1:

Obviously AEF is an isosceles right triangle, so AFM = 45°, i.e. FM bisects AFC.

Take the midpoint n of BC, and the following proves that M and N are the same point, because the bisector of AFC is unique, that is, it proves that Fn also bisects AFC.

Because AFC and ANC are right triangles, on=of=(1 2)ac=oc.

Since on=of, ofn= onf, so ofn=(1 2)(ofn+ onf)=(1 2) nog.

Because oa=of, the same goes for ofa=(1 2) aog.

So afn= ofa- ofn=(1 2)( aog- nog)=(1 2) aon=45° (obviously aon=90°).

So fn also bisects afc, so m is the same point as n, i.e. m is the midpoint of bc, so bm=mc.

Method 2: If you have learned the four-point circle, this problem will be much easier.

Previously we have obtained afm= cfm=45°, so afm= acm, so afcm is conspecific, so cam= cfm=45°, and easy to prove m is the midpoint of bc.

The reason why method 1 is so awkward is to prove that these four points are in a circle.