(2) Take point A as AF⊥AN, intercept AF=AN, connect BF and FM, and work out ∠ 1=∠3 according to the congruence of the same angle, then prove the congruence of △ABF and △AND with "edges and corners", and get BF=DN, ∠ according to the equivalence of corresponding sides in congruent triangles. Then, it is proved that △AFM and △ANM are congruent by "edges and corners", and FM=NM can be obtained according to the equivalence of corresponding sides of congruent triangles, and then it is found that △FBM is a right triangle, and then it can be judged by Pythagorean theorem.
Solution: Solution: (1)∵BM and DN divide the two outer corners of the square equally.
∴∠CBM=∠CDN=45，
∴∠ABM=∠ADN= 135，
∫∠MAN = 45，
∴∠BAM+∠NAD=45，
In △ABM, ∠ Bam +∠ AMB = ∠ MBP = 45,
∴∠NAD=∠AMB，
At △ABM and △NDA,
∠ABM=∠ADN
Nader =∠AMB
∴△ABM∽△NDA，
∴AB/DN=BM/AD，
∴bm&； #8226; DN = AB & amp#8226; AD = a2
(2) The triangle surrounded by 2)BM, DN and MN is a right triangle.
The proof is as follows: As shown in the figure, the intersection point A is AF⊥AN and intercept AF=AN to connect BF and FM.
∫≈ 1+∠BAN = 90，
∠3+∠BAN=90，
∴∠ 1=∠3,
At that sum of △ABF and △,
AB=AD
∠ 1=∠3
AF=AN
∴△ABF≌△AND(SAS)，
∴BF=DN,∠FBA=∠NDA= 135，
∠∠FAN = 90，∠MAN=45，
∴∠ 1+∠2=∠FAM=∠MAN=45，
At △AFM and △ANM,
AF=AN
∠FAM =∠ people
AM=AM
∴△AFM≌△ANM(SAS)，
∴FM=NM，
∴∠fbp= 180-∠FBA = 180- 135 = 45，
∴∠FBP+∠FBM=45 +45 =90，
∴△FB△ is a right triangle,
FB = DN，FM=MN，
The triangle surrounded by BM, DN and MN is a right triangle.
Comments: This topic examines the properties of a square, congruent triangles's judgment and properties, the inverse theorem of Pythagorean theorem, and similar triangles's judgment and properties. The difficulty lies in (2) constructing congruent triangles and right triangle as auxiliary lines.
Hope to adopt, thank you.
Hope to adopt, thank you.