IN ABC, AB BC 10, POINT M N ON BC SUCH THAT MN AM 4, ANGLE MAC ANGLE BAN, FIND THE AREA OF THE TRIAN

Because ab=bc

So bac= c

and mac= ban

So 2 mac + nam = c

and mn=am

So nam= anm

amn= mac+ c

amn=180°-2∠nam

i.e. mac+ c=180°-2 nam

mac+2 mac+ nam=180°-2 nam, so bam= nac= mac+ nam=60° over b as bg am in g, over c as ch am in h in rt abg, ab=10, bag=60°, so bg = 5 root number 3

According to the cosine theorem, we can find BM=2, root number 19, CM=10-2, root number 19, so ch bg=cm BM, we can find CH=5, root number, 3 (10-2 root number, 19) and 2 root number, 19

So the area of the triangle ABC = 1 2 * AM * (BG + CH) = 1 2 * 4 * [5 root number 3 + 5 root number 3 (10-2 root number 19) 2 root number 19] = 2 * 5 root number 3 [1 + (10-2 root number 19) 2 root number 19] = 2 * 5 root number 3 * 10 2 root number 19 = 50 root number 57 19

Because ab=bc, bac= bca

And because bam= can

So 2 dismantle bam+ mac=

Because can+ c= anm

Again, mn=an, we get nam= amn, anm=180°-2 amn, so ban+ c=180°-2

The sum of the two formulas gets:

3 rows of bam+3 gears: man=180°, mac= bam+ man=60°

Solution: Extend bn to AC to e, because AN bisects BAC and BN perpendicular AN, so ban= can, anb= ANE=90, and AN is a common edge.

So abn aen (asa).

So bn=en, ab=ae=14, so ce=ac-ae=ac-ab=19-14=5 and point m is the midpoint of bc, so mn is the median of bc, so mn=ce 2=5 2

Solution: Extend bn to AC to e, because AN bisects BAC and BN perpendicular AN, so ban= can, anb= ANE=90, and AN is a common edge.

So abn aen(asa), so bn=en, ab=ae=14, so ce=ac-ae=ac-ab=19-14=5

And point m is the midpoint of BC, so Mn is the median of BCE, so Mn = CE 2 = 5 2

Solution: Extend bn to AC to e, because AN bisects BAC and BN perpendicular AN, so ban= can, anb= ANE=90, and AN is a common edge.

So abn aen(asa), so bn=en, ab=ae=14, so ce=ac-ae=ac-ab=19-14=5

And point M is the midpoint of BC, so Mn is the median line of BCE, so Mn = Ce 2 = 5 2 Usually be serious in class, in fact, it is very simple.

Method: 1) Make point c with respect to the symmetry point c of ab'to connect to the AC'.

2) Extend PM, alternating AC'Yu d;

3) Intercept an=ad. on AC

Then the point d is the point that is required.

At this time, the perimeter of the triangle PMN is minimal).

In fact, this topic is not an Olympiad, the second and third years of junior high school, the questions that may often be encountered in the mid-term final exam, a little difficult, another netizen's solution, calculation error, the following gives a different solution.

Extend BM to AC on D

AM bisect angle BAC

bam=∠dam

am=amamb=∠amd=90°

amb≌△amd(asa)

ab=ad=8，bm=md

cd=12-8=4

n is the midpoint of BC.

mn = 1 2cd = 2 (median nature).