What is Apollonius Theorem? How useful it is? | Proof with Examples

What is Apollonius Theorem? How useful it is? | Proof with Examples | Step by step explanation

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PreMath

Күн бұрын

Пікірлер: 52

@rishudubey1533 Жыл бұрын

thankyou so much dear professor 😊

@PreMath Жыл бұрын

You are most welcome, Rishu dear

@misterenter-iz7rz Жыл бұрын

Would it simply applying cosine rule? Let a=AB, b=AC, m=AM, n=BM, x=angleAMB, so a^2=m^2+n^2-2mn cos x, b^2=m^2+n^2+2mn cos x, adding them, the result follows.🙂

@soli9mana-soli4953 Жыл бұрын

Great! You only have to add that cos(180°-x)= - cos x

@hanswust6972 Жыл бұрын

That's what I did looking for the fastest solution. Nevertheless, Geometric solutions are quite elegant.

@adrianoparzianello36 Жыл бұрын

The cosine rule would do the job, but it is always good to have more tools in our toolbox, and it makes us think further ahead besides that we don't need to get embroiled with angles and trigo. Math really makes us think out of the box.

@rutvikmore3879 8 ай бұрын

Thank You Sir

@alster724 Жыл бұрын

Wow, the Apollonius Theorem proof and application are exciting to solve

@francklecot7011 Жыл бұрын

with the law of cosines cos(B) = (AB²+(2BM)²-AC²)/ (2 AB * 2 BM) AM² = AB² +BM² - 2 AM.BM.cos(B) AM² = AB² +BM² - 2 AM.BM. [(AB²+(2BM)²-AC²)/ (2 AB * 2 BM)] AM² = AB² +BM² - AB²/2 - 4 BM²/2 + AC²/2 => 2 AM² = 2 AB² + 2 BM² - AB² - 4 BM² + AC² 2 AM² = AB² + AC² - 2 BM² => AB² + AC² = 2 AM² + 2 BM²

@PrithwirajSen-nj6qq 6 ай бұрын

It is a special case of Stewart theorem. Pythagoras theorem is a special case of Ptolemy theorem. In brief it is a proof of a corollary with the help of a corollary

@wackojacko3962 Жыл бұрын

I'm going with a quote from Gottfried Wilhelm von Leibniz..."He who understands Apollonius will admire less the achievements of the FOREMOST men of later times." I love the math too prove this theorem and the Visual of actually putting squares on the edges of the triangle and bisector, and seeing the special case of Pythagoras .🙂

@adrianoparzianello36 Жыл бұрын

This quote is beautiful and says all. Only math can display so much beauty. This makes me remind another one " I think therefore I am" we have a short time but it's worth it if we live our life to the full. Take care Amigo

@neetusharma5403 2 ай бұрын

Thank you so much sir now I understand this theorom clearly 😊

@kazisorifulhoque2340 Жыл бұрын

A lot of thanks.i have knew a new formula appolonias theorem.

@SanjayTelangre Жыл бұрын

Thank you so much sir

@HappyFamilyOnline Жыл бұрын

Great 👍 Thanks for sharing😊😊

@hichamitani6433 Жыл бұрын

Excellent prof

@SanjayTelangre Жыл бұрын

Thank you so much sir but very hard trip

@quigonkenny 9 ай бұрын

So is 11-12-17-13 an Apollonius Quadruple?

@therock-jn5qn Жыл бұрын

Thanks a lot It really helped

@KAvi_YA666 Жыл бұрын

Thanks for video.Good luck sir!!!!!!!!!!!

@SAHIRVLOGCLP Жыл бұрын

Thank you so much Proffesor....

@gymmonster8137 Жыл бұрын

Thank you for the helpful explanation to prove the theorem. Great effort. But I have to note, the dimensions of the triangle cannot represent an acute triangle, but rather an obtuse triangle. This caused me difficulty in solving the example first before watching the video, using the law of cosines. .... I also tried the Heron's Formula to get the area of the triangle, then get h, then get the median m.. But it was confusing because the results didn't match because the dimensions of the triangle are not correct for an acute triangle..

@ybodoN Жыл бұрын

Heron's formula works this way: let AM = x so the semiperimeter of △ABM is (11 + 6 + x) / 2 = (17 + x) / 2. Half the area of △ABC is √1080 = √((((17 + x) / 2) (((17 + x) / 2) − 6) (((17 + x) / 2) − x) (((17 + x) / 2) − 11))). Simplify to 17280 = (17 + x) (5 + x) (17 − x) (x − 5) leading to the biquadratic equation x⁴ − 314 x² + 24505 = 0. Solutions are x = −13, x = 13, x = −√145 or x = √145. With Appolonius theorem we directly get x = −13 or x = 13.

@osumanuabubakar9557 Жыл бұрын

Great theorem!

@aminex3519 Жыл бұрын

Excellent❤ as always thank you

@Abby-hi4sf Жыл бұрын

Lovely!!!

@peterlblystone2326 Жыл бұрын

I used trigonometry, mainly the law of cosines, and got the same answer.

@walcholjacob4259 Жыл бұрын

Wonderful.

@vacuumcarexpo Жыл бұрын

I thought the finder of this theorem is Pappus, but I knew Apollonius is the real finder thanks to this video.

@lraoux Жыл бұрын

I tried solving this initially but ran into a problem- can you explain why? (6+x)^2+h^2=17^2=289= 36+12x+x^2+h^2 (6-x)^2+h^2=11^2=121= 36-12x+x^2+h^2 Subtract the 2 equations and get 168=24x -> x=7 But x cannot =7! 6+7=13 > 12. Why does this happen?

@mohamadtaufik5770 Жыл бұрын

I think the question is wrong, line BD isn't negative

@WernHerr Жыл бұрын

Calculate the angle (ABC) and get 95,21, so h ist outside ABC and x is indeed 7.

@mohamadtaufik5770 Жыл бұрын

From the skecth given, it looks the angle ABC is lower than 90 degree, so the sketch must be revised

@mohamadtaufik5770 Жыл бұрын

If the line AM = 13 and m=a-x, 11-m^2=13^2-(6-m)^2 m=-1, therefore the question is wrong because BD isn't negative

@WernHerr Жыл бұрын

The triangle is an obtuse triangle!

@mohamadtaufik5770 Жыл бұрын

I've recalculated, regardless the triangle is obtuse or acute, if the value of x is positive meaning the triangle is acute triangle, and if the value of x is negative meaning the triangle is obtuse triangle, therefore the line AB is moving to the left so that AB and BC creates angle above 90°. Using phytagorean theorem, AM^2=h^2+BM^2 AM^2= (2V30)^2+(7)^2=120+49 --> AM=13

@lzuluaga6064 Жыл бұрын

Me gustó. I liked.

@MuhammadGulzar-i6v 9 ай бұрын

Wow ❤ Ow

@MukeshSingh-hh5je Жыл бұрын

Here we also use Stewart therom

@ybodoN Жыл бұрын

Apollonius' theorem is a special case of Stewart's theorem. Also, it is almost 2000 years older…

@aminex3519 Жыл бұрын

I watch only your math videos

@johnyriosrosales7674 Жыл бұрын

Usando teorema del coseno sale muy fácil de hecho nisiquiera es necesario calcular ángulos es simple: C^2=A^2+B^2-2ABCcosc C=17 A=11 B=12 17^2=11^2+12^2-2×11×12cosc 289=121+144-264cosc 289=265-264cosc 264cosc=265-289 264cosc=-24 cosc=-24/264 cosc=-1/11 Luego aplicamos de vuelta el teorema en este caso C es la longitud que vamos a calcular y A=6 B=11 así que ahora si reemplazando C^2=6^2+11^2-2×6×11×cosc Recordemos que cosc=-1/11 C^2=36+121-2×6×11×(-1/11) C^2=157+12 C^2=169 Y como recordamos que C es una longitud entonces tomamos el valor positivo dando que la longitud que nos mide el problema es 13u

@anastasentezilyayo5279 Жыл бұрын

J'ai d'abord calculé x et h pour trouver d d'après le théorème de Pythagore. Mais j'ai trouvé que x =7. Ce qui est impossible. A mon avis les dimensions du triangle ne sont pas correctes.