Who has the questions in the first volume of eighth grade mathematics?

(1) In the triangle ABC, P is a point above BC, PR⊥AB is in R, PS⊥AC is in S, AQ=PQ, PR=PS, which proves that (1)AS=AR(2)PQ is parallel to AR.

Proof: even AP

Because PR⊥AB is in R, PS⊥AC is in S, PR=PS,

So AP is the bisector of the angle BAC,

So the triangle ARP is equal to the triangle AQP,

So AR=AS,

(2) In △ABC, AB=AC, D and M are the midpoint of AC and BC respectively, E is a point on the extension line of BC, and CE = 1/2bc proves that DB=DE.

To prove that two sides are equal, it is often proved by proving that the lines of two triangles are equal.

In the problem, triangles that are more likely to be congruent are △DMB and △DCE.

At this point, the excavation conditions are known.

AB=AC, m is the midpoint of BC, then BM=CM,AM⊥BC,∠ABC=∠ACB,

And d is the midpoint of AC, so DM//AB, and DM= 1/2AB, ∠DMC=∠ABC.

But we just need to prove that the triangle is congruent, and we don't need so many conditions.

You need to know what theorem can prove that two triangles are congruent. Corner, corner and edge are all ok, depending on the actual situation.

Proof: Because AB=AC, M is the midpoint of BC.

So BM=CM,AM⊥BC,∠ABC=∠ACB.

And CE= 1/2BC, so BM=CE proves that a congruent edge comes out.

D and m are the midpoint of AC and BC, respectively.

Therefore, DM//AB, ∠DMC=∠ABC=∠ACB. Equiangular.

So DM=DC because ∠DMC=∠ACB, △DMC is an isosceles triangle.

∠DMB=∠DCE because ∠DMC=∠ACB, they are the complementary angles of ∠DMB∠DCE respectively.

So, in △DMB and △DCE,

BM=CE, ∠DMB=∠DCE, DM=DC edge theorem

So △ dmb △ DCE, so DB=DE.

(3) In △ABC, AB=AC, D is a point outside △ ABC, and ∠ Abd = ∠ ACD = 60, and the verification CD=AB-BD.

Proof: extend CD to e and make ACE an equilateral triangle.

Therefore, AC=AE=CE, and because AC=AB, AB=AE, that is, angle ABE= angle AEB.

Since angle ABD= angle AED=60 degrees and angle DBE= angle DEB, BD=DE.

So AB-BD=CE-DE=CD

(4) The length of the bottom of the isosceles triangle is 10cm, and the center line of the waist is drawn from one end of the bottom. The circumference of a triangle is divided into two parts, one of which is 4cm longer than the other, which is the waist length of an isosceles triangle.

Suppose the length of the waist is a.

A+a *1/2 =10+a *1/2+4 or a+a *1/2+4 =10+a *1/2.

A= 14 A=6

Answer: The waist length of an isosceles triangle is 6 cm or 14 cm.

(5) As shown in the figure, in △ABC, AB=AC, point E is on AB, point D is on the extension line of AC, CD=EB, and ED and BC intersect at point M. Verification: EM=DM.

Proof: do EF‖AD to BC and f through point e,

So ∠FEM=∠D, ∠ACB=∠EFB.

AB = AC

∴∠B=∠ACB

∴∠B=∠EFB

∴EB=EF

CD=EB

∴EB=CD

In △EFM and △CDM,

∠CMD=∠EMF，

∠FEM=∠D，

EB=CD