A difficult geometry problem, the master helps

In this text, (bac) (acd) is used instead of angangle bac and acd

s(abcd) = s(abc) +s(acd)= 1/2 * ab*ac*sin(bac) +1/2 * ac*cd*sin(acd)

1/2 * ac*(ab*sin(bac) +cd*sin(acd))

1/2 * ac*(ab+cd)

1/2 * ac*(16-ac)

1/2 * ac +16 - ac)/2)^2 = 32

The equal sign holds if and only if ac= 8 sin(bac) = sin(acd) = 1.

There are 7 such quadrilaterals.

It can be divided into four groups, each of which has congruent quadrilaterals, but the letter numbers are different.

ab =1,cd =7, ac =8, sin(bac) = sin(acd) = 1

ab =7,cd =1, ac =8, sin(bac) = sin(acd) = 1

ab =2,cd =6, ac =8, sin(bac) = sin(acd) = 1

ab =6,cd =2, ac =8, sin(bac) = sin(acd) = 1

ab =3,cd =5, ac =8, sin(bac) = sin(acd) = 1

ab =5,cd =3, ac =8, sin(bac) = sin(acd) = 1

ab =4,cd =4, ac =8, sin(bac) = sin(acd) = 1

No. In planar geometry, the area of the graphic circle with the same perimeter is the largest.

It is only required that the area of the circle with a circumference of 16 is not 32, which proves that there is no such figure.

The area of the circle with a circumference of 16.

Pi*< (16 Pi) 2>

3^n - 1）/ 2

The number of white is increased to the number of black triangles in the previous graph, so that the nth diagram is recursively summed when the first n terms of the series with the initial term of 1 and the ratio of male pants are 3.

Theorem 2: The bisector of an angle of a triangle divides the two line segments formed by the opposite side of the opposite side in proportion to the correspondence of the two sides of the angle, e.g., in ABC, CE bisects BAC, then BC:AC=BE:AE because BE=AE

So be:ae=1

This is because the quantitative relationship between AC and BC is uncertain.

So ac:bc=(we don't know) from your graph, it should be that ac:bc is not equal to 1, because ac:bc is not equal to ae:be

So it does not conform to the angular bisector theorem.

So CE is not an angular bisector.

For example, AC=BC, CE is the angular bisector!!

Because |af2| =2a - af1|,|bf2| =2a - bf1|

So |af1|+|bf2|+|ab| =4a-|af1|-|bf1|+|ab|

ab|-|af1|-|bf1|<=0

And the three good ones are sent to a friend when the straight line,|ab|-|af1|-|bf1|=0, the maximum perimeter is 4a

Through a as the vertical axis of AP, through B as the vertical axis of BQ, ap=sinx*af1, ab=sinx*f1, area = 1 2(ap*f1f2+bq*f1f2)1 2 f1f2*(ap+bq).

1/2 f1f2*sinx(af1+bf1)1/2 f1f2*sinx*ab

Maximum when x = 90 degrees.

That is, when AB stares at the bridge on the vertical X-axis.

Area max = 1 2 f1f2 * ab

Use the method of counter-evidence (or the same law).

a=∠a，ad=ae

To prove ADC AEB, one only needs to prove ab=ac assuming ab≠ac, then abac

a=∠a，ad=ae，ac'=ab

adc‘≌△aeb

ac'd=∠abe

Plus do'b=∠eo'c’，db=ab-ad=ac'-ae=ec'

do'b≌△eo'c’

o'b=o'c'

o'b-ob=o'c'-oc, i.e. o'o=o'c'-oc∴o'o+oc=o'c'

But AED is an acute angle.

dec is an obtuse angle, o'c'C> dec is also an obtuse angle o'c'c>∠o'cc'

o'o+oc>o'c>o'c'Contradiction!

In the same way, when AB>AC can also push out contradictions.

If the assumption is not true, ab=ac

adc≌△aeb

Dude, there is a lack of conditions, such as b= c, or ado= aeo, etc., and then it is well proven

Certificate Abe ACD:

Proof: Because ADC= AEB, and AD=AE, A is common, then Abe ACD is obtained from the corners. Other conclusions are based on this.

Connect the BC

The extension line of the AO crosses BC at point F

Because ob=oc

So obc= ocb

of=of, so obf ocf, so the point f is the midpoint of bc, and af is called bisector of boc, so af is also the angle bisector of bac.

AFB AFC can be obtained

ac=ab, and because ad=ae, dac=eab

So ADC AEB

Report|Share|2016-10-31 23:53The man who chases the wind |Viewed 30 times Bounty: 50

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I don't know how to do it, but I can tell you the answer by brute force, 6°

You can try it with trigonometric functions.