Proof: Basic Proportionality Theorem for Triangles

Proof: Basic Proportionality Theorem for Triangles | Geometry

Рет қаралды 3,594

3 жыл бұрын

If a line is drawn parallel to a side in a triangle, so that it intersects the other two sides in two distinct points, then those two sides are cut in the same ratio. This is the basic proportionality theorem for triangles, very related to similar triangles. It is also called the intercept theorem or Thale's theorem.
We'll prove this theorem in today's geometry lesson, making use of the fact that triangles with the same base between the same parallels have the same areas. This will allow us to construct equal ratios of areas of different triangles. Note, in the video, triangle ADE is similar to triangle EBC. This is because the triangles have equal angles (they share angle A, then B and D are corresponding angles on parallel lines, and similarly for E and C). Thus, not only is AD to DB as AE is to EC, but these ratios are also the same as the ratio of DE to BC.
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I hope you find this video helpful, and be sure to ask any questions down in the comments!
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Пікірлер: 24

@suggi6860 3 жыл бұрын

Very much thank you!!.....I was so confused in this proof ...You are amazing

@WrathofMath 3 жыл бұрын

You're very welcome! I'm glad to hear it helped and let me know if you ever have any video requests!

@AbhishekRai-qo2fj 2 жыл бұрын

Love from India sir

@nolansharma9891 2 жыл бұрын

Boards kaisa gya bhai

@zaini7255 Жыл бұрын

thank you sm!!!!!!!! the explanation was amazing and rlly easy to understand

@WrathofMath Жыл бұрын

Glad to help!

@SO111OO11 10 ай бұрын

You are a really good teacher Sir! I just wrote my board exam and this is EXACTLY what came in the exam. I remembered it so well i completed the whole proof in 5 mins 😂😂. Thank you so much for the video. I will definitely be keen to check on future lessons!

@BornToThinkExpressively 2 ай бұрын

Beautiful, just beautiful 🥲

@WrathofMath 2 ай бұрын

Thank you!

@Gaelrenaultdu06 3 жыл бұрын

Hello, could you prove the 3rd equality ? DE/BC = AE/EC ?

@ganeshkanse1848 3 жыл бұрын

Its gone very helpfull to me thanks 🙏

@WrathofMath 3 жыл бұрын

So glad it was helpful! Thanks for watching!

@jonathanmartinbenitez7427 Жыл бұрын

Thanks Lot !!!

@WrathofMath Жыл бұрын

My pleasure!

@krishorepalanati7980 3 жыл бұрын

Thanks sir

@WrathofMath 3 жыл бұрын

No problem, thanks for watching and for the request!

@migfed 9 ай бұрын

Beautiful

@WrathofMath 9 ай бұрын

Thank you!

@rssl5500 3 жыл бұрын

Hello I have a question I think the proof has a problem you drew altitude from point E to AD but what if there is an acute angle that is an gone bigger than 90 degrees then you can’t draw inner altitudes so this assumption is wrong

@WrathofMath 3 жыл бұрын

Thanks for watching and for your question! I just looked at the proof briefly to see your question, so I could be wrong - but I'm pretty sure we never use the assumption that the altitude EP falls inside of the triangle. What's important is that it intersects line AB, it's the distance from that line to point E, and this is the height of two of our triangles. By the line AB I mean if we were to continue segment AB infinitely in either direction to create a line. If angle A is obtuse, then like you say the altitude from E will fall outside of the triangle, but that doesn't break anything in the proof, it would just make us have a slightly uglier diagram. Hope that helps!