thankyou so much dear professor 😊

@PreMath

Жыл бұрын

You are most welcome, Rishu dear

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

Great 👍 Thanks for sharing😊😊

Thank You Sir

Excellent❤ as always thank you

Thank you so much Proffesor....

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

Thank you so much sir

Excellent prof

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

Thanks a lot It really helped

Lovely!!!

Great theorem!

Wonderful.

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

8 ай 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.

Me gustó. I liked.

Thank you so much sir but very hard trip

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

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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

8 ай 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

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

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

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

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

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.

I watch only your math videos

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.

1:53

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

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…

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

More math question on my channel

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

rad145...ho sbagliato i calcoli.. Ah ah

Interesting, but not sure it that useful... Hey listen to me pretending I know some stuff 😂😂😂 Thanks for another geometry gem

טעות בשרטוט הYOU TUBE לפי הגאו גברה וספר הנדסה חלק ב' זווית B קהה

I don't understand anything u are teaching in this video

@user-ti8di3xh4d

5 ай бұрын

kzread.info/dash/bejne/gol8zchpibG8ZNY.html

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