In ABC, BAC 90, AB AC, M is the midpoint of the BC edge, and MN BC intersects AC at the point N

1. Because a=90

So b+ c=90

Also because of MN BC, MQ MP

So C+ CNM=90, BMP+ PMN=90, PMN+ NMQ=90, so B= CNM

bmp=∠nmq

So: PBM qnm

Therefore, there is nq bp=nm bm---1).

2、∠abc=60°，ab=4 √3

c=90-60=30

ac=√3ab=12

bm=mc=1/2bc=ab=4 √3

nm=1/2nc=√3/3mc=4

nc=8 with (1) formula.

nq==√3/3bp

Because p velocity vp = 3

So q velocity vq = 1

The area of the APQ is S

ap=ab-bp=4 √3-t√3

aq=ac-nc+nq=12-8+t×1=4+ts△apq=ap×aq/2=(4 √3-t√3)(4+t)/2s△apq=8√3-(√3/2)t∧2

0 In the right angle abc, abc=0°, ab= cm, then bc= cm, ac=cmBy m angle bac = 0 °, ab< ac, m is the midpoint of the bc side, and mn bc crosses ac at the point 0 0 -- Zhao Di S< P>

as he ab

AMB is similar to HMA

ma/mb=ah/ab

ab = 2 root number slow blind elimination 2, ma = 1, mb = root number 5ah = (2 root number 10) 5

ah=hd=(2 root number 10) 5

he⊥abae=ed=1/2ad

cbm=∠abm=1/2∠cba

Omit a few steps in the middle to disturb the knowledge. Prove yourself to the next step.

hab=∠cbm

cos cbm=cb mb=(2 root number 5) 5cos god this had=(2 root number 5) 5

ae = (4 root number 2) 5

ad=(8 root number 2) 5

Let's prove it for yourself with the following two.

Root No. 10) 5).

Root No. 2 2).

AHD is an isosceles triangle, which two sides are equal? ah=ad or ah=dh or ad=dh? Three cases.

The question is wrong, how can the hypotenuse of a right triangle be the same length as the right angle?

Proof: Connect AM

an²=am²-mn²

bn²=bm²-mn²

an -bn nucleus tease = am -bm

Because m is the midpoint to change the finch to sell, BM=cm

So an -bn = am -bm = am -cm = ac

Idea: Transfer the three sides to the same triangle, and then use the trilateral relationship of the triangle to solve it.

Proof: 1Extend no to p so that no=op and link bp

2.The EPO of the triangle is equal to the triangle CNO, so nc=bp3In the triangle MOP and the triangle mon, PO = on, angular MOP = angular mon = 90 degrees, MO = MO

So the triangle MOP is all equal to the triangle mon, so mp=mn4In a triangle BMP, BM+BP is greater than MP

5.So BM+CN>mn

There should be a lot of methods, this is a conventional solution, I hope it helps.

Extend the mo, cross the parallel line of the point c to do ab cd the extension line of the mo at the point d, connect the dn

Since O is the midpoint of BC, CD BM, which proves BMO CDO, we can get BM=CD, MO=OD, and ON OM, so MN=DN, in CDN, CN+CD DN is BM+CN>MN

Stealth is visible to the visibility, hello :

The proof is as follows: the connection am is an isosceles right triangle because abc is 90° in abc, ab ac, a is aware of 90°, and c is 45° because m is the midpoint of bc.

So am cm, am bc, bam cam 45° so bam c

Because df ab, de ac, a 90°

So the quadrilateral afde is rectangular.

So af de

Because the triangle CDE is an isosceles right triangle.

So ce de

So af ce

So AFM Royal Lead CEM (SAS).

So em ef, good amf cme

So amf ame cme ame is fme amc 90°

So mef is an isosceles right triangle.

It must be a right triangle. But if it's not an isosceles right triangle, I don't know the Chaqing proof. Because abc is a right-angled imaginary triangle.

So a=90°

Because ME is perpendicular to AC

So mec=90°

mec=∠a

So ME is parallel to AB

So lose the reputation of holding emc= b

And because DF is perpendicular to AB

So b+ fdb=90°

So emc+ fdb=90°

And because FME+EMC+FDB=180

So fme=90°

Therefore, --- is a right triangle.

1 All sin mch is mh cm, where cm=1, bc=2, so mb = root number 5, triangle mch is similar to triangle mbc (the two angles are equal), so there is: mh mc = mc mb can solve mh mc = 1 root number 5, i.e. sin mch = root number 5 5

Extend AD to F to connect FC so that FC is perpendicular AC = AB angle BAC = angle ACF angle ABE = angle CAE =