This proof is very nice, it uses a very simple argument for the ratio of the sides and for the median, that's what's so cool about geometry I like it more than the coordinate proof
Remember, people, perpendicular bisectors meet at the circumcentre ( centre of an excribed circle), altitudes meet ( are concurrent) at the orthocentre, and medians meet at the centroid.
Vlad Chira I'm thinking it's handover if any triangle inscribed or describes a circle. Again, when is it appropriate in real life, though? Maybe reengineering problems?
I cant help but to fail to see that concurrency of the medians cannot be "proven" in the sense of the word. To me, its more like an instance of a "magical miracle" that all three of these lines meet at one, unique point. Question: Is this theorem, being a specific case of the Ceva Theorem itself, named after a famous mathematician? Just seems like it would to make it easier to memorize.
A=4B, A+20 = 2(B+20), subtitute, A+20= 2(B+20) 4B+20=2B+40 4B-2B=40-20 2B=20 B=20/2 B = 10 So we know when 1980 age of son is 10 years old, so he will birth in 1970
Since both triangles share that same interior angle at C and the legs that form it are both proportional (by a factor of 2), by SAS (side angle side) they must be similar.
To prove two triangles are similar, it is sufficient to show that the ratios of two pairs of corresponding sides are proportional and the angles they include are congruent.
This proof is very nice, it uses a very simple argument for the ratio of the sides and for the median, that's what's so cool about geometry
I like it more than the coordinate proof
Proud for proving it myself
Remember, people, perpendicular bisectors meet at the circumcentre ( centre of an excribed circle), altitudes meet ( are concurrent) at the orthocentre, and medians meet at the centroid.
John M Apart from centroid, remembering the intersection point is kinda useless (exept maybe from some radiuses in the triangle)
Vlad Chira I'm thinking it's handover if any triangle inscribed or describes a circle. Again, when is it appropriate in real life, though? Maybe reengineering problems?
Those points make the Euler line
Very convincing demo! Thanks.
You are great teacher
I wanted to ask a question. Would that median also be in a ratio 1:2. Coz it trisects right?
I cant help but to fail to see that concurrency of the medians cannot be "proven" in the sense of the word. To me, its more like an instance of a "magical miracle" that all three of these lines meet at one, unique point.
Question: Is this theorem, being a specific case of the Ceva Theorem itself, named after a famous mathematician? Just seems like it would to make it easier to memorize.
thanks!!!!!!!!!!!!!!
sir I am not finding the solution of an question?
in 1980 if the age of father is 4 times than son.In 2000 father age becomes 2 times than son.find sons birth year?
Trial and error gives you 1970, but I don’t know the algebraic method
A=4B, A+20 = 2(B+20), subtitute,
A+20= 2(B+20)
4B+20=2B+40
4B-2B=40-20
2B=20
B=20/2
B = 10
So we know when 1980 age of son is 10 years old, so he will birth in 1970
thank you
@@JJclassofLearning
Is this a grade school problem??
so you just criticized the proof, didn't show it by yourself in whole 8 minutes and titled the video: Proof via similar triangles. Nice one!
thanks!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Actually, it's not clear for me why is ABC similar to DCE
Since both triangles share that same interior angle at C and the legs that form it are both proportional (by a factor of 2), by SAS (side angle side) they must be similar.
To prove two triangles are similar, it is sufficient to show that the ratios of two pairs of corresponding sides are proportional and the angles they include are congruent.
I like how you guys are still answering this 2 years old question :D
just use midpoint theorem for the first part then prove ABP similar to EDP ez
O cara da frente ao inves de prestar atençao no professor, preferiu copiar a lousa
Renan os caras tentam decorar, por isso que não aprendem