Geometric discussion

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180(n-2)°

I will post a prove soon.

This proves that the interior angles of a triangle add up 180°

I assume that an n-sided polygon can always be divided into n-2 triangles (This mean the sum of interior angles is 180(n-2)°), but it hard to visualize. Especially polygons with some interior angle greater than 180°. So I will prove it more clearly by induction.

Let p(n) represent “the sum of the interior angle of an n-sided polygon equal to 180(n-2)°” and n=k

Base case: for k=3; I have proved above that the sum is 180° = 180(3-2)° so p(3) is true

induction step: assume p(k) is true, we will prove p(k+1) is also true.

Consider any (k+1)-sided polygon. Choose one angle that is less than 180° in that polygon, Then draw a diagonal with two vertex adjacent to the selected angle. Now that diagonal divides the polygon into 2 parts. Is a triangle and k-sided polygon. So the sum of interior angles is equal to 180(k-2)°+180° = 180(k-2+1)° = 180((k+1)-2)°. So p(k+1) is true

By induction, we can concluded that p(n) is true for n=3,4,5,…

Problem 1:

ΔABC is any triangle, prove AB=AC if and only if ∠B=∠C.

HINT: Let D is a point on BC that causes AD⟂BC.

Theory 1.1

If two right triangles have the same length of the hypotenuse and one cathetus, then both triangles are congruent (right-side-side).

Prove Thale’s theorem

Solution: Let D is a point on BC that causes AD⟂BC.

[->]

1. AB=CD (information from question)
2. AD=AD (common side)
3. ∠ADB=∠ADC=90°

So ΔADB ≡ ΔADC and ∠B=∠C

[<-]

1. ∠ABD=∠ACD (information from question)
2. ∠ADB=∠ADC=90°
3. AD=AD (commom side)

So ΔADB ≡ ΔADC and AB=AC

Let A,B,C be different points on circle O. If AB is diameter of circle O, prove that ∠ACB=90°.

Solution: draw a straight line OC

• ΔAOC, ΔBOC isosceles triangle because OA=OC=OB (circle radius)
• ∠OCA=(180°-∠AOC)/2 (interior angles of ΔAOC)
• ∠OCB=(180°-∠BOC)/2 (interior angles of ΔBOC)

So ∠ACB = ∠OCA+∠OCB = (180°-∠AOC)/2 + (180°-∠BOC)/2 = [360°-(∠AOC+∠BOC)]/2 = [360°-180°]/2 = 90°

Problem 2:

Find the area of ΔAEF.

I didnt have any paper to work this out on, so I did it all in my notes along with my phone calculator. Also, I don’t really know the names of the theorems or stuff but I do know what I’m doing. Here’s the answer with the solution. 

AD is 6 cm

Point E is the midpoint of AD

Angle BCD and angle BAC is 45 degrees

Triangle BAE is a right triangle

By the Pythagorean theorem, BE is square root of five.

BA is 6cm

Angle CBA equals angle AEB

tan(AEB)=2

AEB is 63.43494882292201 degrees

EAF is 45 degrees

EFA is 71.56505117707799 degrees.

Law of sines part

3/sin(71.56505117707799 degrees)=x/sin(63.43494882292201 degrees)

x=3sin(63.43494882292201 degrees)/sin(71.56505117707799 degrees)

x= 2.82842712474619 cm = AF

repeat this for y

3/sin(71.56505117707799 degrees)=y/sin(45 degrees)

y=3sin(45 degrees)/sin(71.56505117707799 degrees)

y= 2.23606797749979 cm = EF

(2.23606797749979 + 2.82842712474619 + 3)/2= 4.03224755112

sqrt(4.03224755112(4.03224755112-2.23606797749979)(4.03224755112-2.82842712474619)(4.03224755112-3))=2.99999999999999

Therefore, the area of triangle AEF is 2.99999999999999 cm squared. Or you can round it up to three lol.

That right! here is my solution.

ΔAEF is similar to ΔCBF because ∠EAF=∠BCF, ∠AEF=∠CBF (alternate angle) and ∠AFE=∠CFB (opposite angle).

Since AE:BC=3:6=1:2, So height ratio of ΔAEF:ΔCBF=1:2

Let The height of ΔAEF, ΔCBF equal to h and 2h respectively, so that h+2h=6 -> h=2

Therefore the area of ΔAEF = 0.5×AE×h = 0.5×3×2 = 3 cm²

Problem 3:

Find the approximate height of the building.

I’ll post a solution soon, but it’s approximately 26 meters

Absolutely correct!

Problem 4:

The ship is sailing in the ocean. The boatman was 20 meters above the water surface and could see the horizon 16,000 meters away.

Estimate the length of the earth's radius from this situation.

Exellent method!

A little: 16,000² = 256,000,000. So the equation should be 256,000,000 = 400+40r.

The approximate value of r is 6.4×10⁶ m.

Problem 5:

Find the ratio of the area of ​​equilateral hexagon and equilateral triangle inscribed in circles of equal radius length.

Problem 6:

Given lines AB and CD are parallel, There are M, N as points between the two lines such that ∠ABM=25°, ∠DCN=45°, and ∠BMN=90°.

Find the size of ∠CNM.

Solution:

Therefore, the area ratio is 1:2

Interesting, @thecosmos999 wrote this same question on the public forum. It was the first thread I comments to him. And I became a member of this club.