There are six answers to the math problems and questions in Grade Two.

1. It is known that ABCD is a square, diagonal AC, BD and O intersect, quadrilateral AEFC is a diamond, EH⊥AC, vertical foot H. Verification: EH= 1/2FC.

Solution: EH= 1/2FC

The reason for this is the following：

ACFE is a diamond. Ac parallel EF, AC=FC

In the square ABCD == >; Ac vertical BD, AC =BD

Because EH⊥AC

So EH is parallel to BD,

Ac parallel EF (certification),

So DOHE is a parallelogram.

So EH=OD= 1/2AC.

So EH= 1/2FC.

2. Take Rt△AB, two right-angled sides AB of AC as a square, and the diagonal vertices D.E of ABDM ABDM, ACEN, ∠ BAM and ∠ CAN are vertical lines DF and EG, respectively, and the vertical feet are F and G. 。

Verification: (1) BC = df+eg; (2)S△ABC=S△FBD+S△CEG

The first small question:

Triangle ABC, BDF and CEG are similar, right? Two proportional coefficients x and y are introduced.

x=DB/BC=AB/BC

y=CE/BC=AC/BC

That is, X and Y are the similarity ratios of these similar triangles.

Ok, so DF=x*AB, EG=y*AC.

DF+EG = x * AB+y * AC =(AB * AB+AC * AC)/BC = BC * BC/BC = BC

The second small question:

Similarly, DF*BF=x*AB*x*AC.

CG*EG=y*AB*y*AC

Add the above two formulas, df * BF+CG * eg = (x * x+y * y) * ab * AC = ab * AC.

The left and right sides of the above formula are twice the proof conclusion (divided by 2 is the area)

3. The points M and N are the sides BC of the square ABCD and the points on the CD. It is known that the circumference of △MCN is equal to half of the circumference of square ABCD, so find the degree of ∠MAN.

4. In the ABCD square, point Q is on CD, DQ=CQ, point P is on BC, and AP=CD+CP. Proof: AQ divides equally ∠DAP.

Because the circumference of the triangle MCN is half of the square, so:

MC+CN+NM=BC+CD=BM+MC+CN+ND。

So BM+ND=MN.

Fix point A on triangle ABM, and rotate the rest 90 degrees counterclockwise.

In this way, the original AB edge and AD edge coincide.

Assuming that the origin m is now at e, then:

From the nature of rotation, we can know that:

Triangle ABM is equal to triangle ADE.

In Eminem Triangle and AEN Triangle:

MN=BM+ND=DE+ND=NE，

Ann is on the edge of the public,

AM=AE。

So triangle AMN and triangle AEN are congruent!

So Angle MAN = Angle Man = Angle Dan+Angle Day = Angle Dan+Angle Bam.

but

Angle MAN+ angle DAN+ angle BAM= angle BAD=90 degrees, so:

90 degrees =2* angle, male.

So: Angle male =45 degrees.

5. At △ABC, ∠ C = 90 AC = BC, let any point on AB be E in F, PE⊥BC in PF ⊥ AC, and M be the midpoint of AB. It is proved that △MEF is an isosceles triangle.

The same topic:

In Rt△ABC, AB=AC, ∠A=90 degrees, D is any point on BC, DF⊥AB,DE⊥AC,M is the midpoint of BC, so what triangle is △MEF?

Analysis: M is the midpoint of the bottom of isosceles △ABC, so we might as well connect AM and apply the property theorem of isosceles triangle "three lines in one". Conclusion: △MEF is an isosceles right triangle.

Proof: link AM

∫∠BAC = 90, AB=AC, and m is the midpoint of BC.

∴AM

=BM，∠BAM=∠CAM=45 ,AM⊥BC

∵DF⊥AB,DE⊥AC,∠BAC=90

∴ Quadrilateral is a rectangle, ∴DF=AE.

∵DF⊥AB,∠B=45 ,∴∠FDB=45 =∠B

∴BF=DF,∴BF=AE

At △BFM and △AEM,

∴FM=EM,∠BMF=∠AME

∴AM⊥BC,∴∠BMF+∠AMF=90

∴∠AME+∠AMF=∠EMF=90