Angle bisector theorem proof | Special properties and parts of triangles | Geometry | Khan Academy

[@nirvairsingh3416]

Have to admit it your way of teaching is simple and cool . I wish u could be my maths teacher keep up the good work man🔥🔥🔥🔥

[@evanpalmer3480]

cool maths, thanks khan

[@oo88oo]

Wow. Great. I wasn’t coming up with that one though I tried and tried. And I had to watch the video to the end and restart it to finally get the proof. Now of course it looks almost trivial once I construct and consider that under-triangle which is similar. Thanks!!

[@geraldvillanueva2997]

thank you!!!! i have more idea for my report!!!its so very helpful for me THANK'S

[@168grhp]

Most helpful!!!

[@selvakumarteacher9798]

That's awesome sir 👏👏👏👏❤

[@kaushikdr]

Basically, can you say that since 1) BCA ~ FDC (same angles) and BDC = FDC (SAS), that BCA ~ BDC?

[@vancedforU]

how did you get BDC=FDC?

[@kelvin7743]

​@@vancedforU Triangle Congurence.

[@anushanaidoo3062]

I am confused

[@aayushkulkarni8939]

Why

[@Md.abdullah1811]

Which part?

[@AqifSaifi]

The subs are at 6.99M , Nice!

[@sanjaydubey6172]

Can't understand what is said

[@holyshit922]

This theorem can be used to derive u sub for R(x,sqrt(ax^2+bx+c)) integrals after completing the square

[@j_apollo9863]

when

[@zack_120]

1:06 - Talking about this theorem the triangle used for demonstration needs to be obviously different from an equilateral one like this one used here where the side-to-side ratios ARE visually same thus no need of proving.

[@antoniopedrofalcaolopesmor6095]

Indeed, for an equilateral triangle you would know beforehand the result is valid just by symmetry. Anyways, the reasoning in the video is valid for any triangle, equilateral triangles included. I agree with you though, for the sake of the general case it would have been preferable to use a "not so equilateral" triangle, I mean whose side lengths are more distinct.

[@sajjansingh3418]

Thank you Sir

[@kaylab8388]

I love ur videos

[@sanjeevsrivastava3389]

Man that was so smooth!!!

[@vagrambagratunian2876]

Excellent!

[@svnvoo]

can we also set the proportion as AB/BC = AD/CD?

[@kailashmuralidharan5849]

same q

[@antoniopedrofalcaolopesmor6095]

Yes, that is equivalent. Just multiply both sides of the identity in the video by AD/BC and you'll get your identity.

[@krishnakrishna-t3e]

yes

[@aasthajain6212]

Thank you khan academy for the proof❤

[@Simba.Luna.]

Great!

[@zack_120]

This triangle used is almost isosceles . If cut line BC very short, say

[@antoniopedrofalcaolopesmor6095]

No matter the triangle you start with, the reasoning in the video is always valid.

[@oo88oo]

It’s hard to construct the drawing without some sides looking close in length. But it can be done. I used GeoGebra. Or you can use a big piece of paper. You need a mix of very acute and very obtuse angles to see the proof clearly.

[@martinandani]

THAHK YOU DUDE

[@paulwang7229]

in fact you can use the ratio of areas of triangle ABD and CBD. This ratio equals to both AD/CD and AB/BC. In the first case they have the same height (the distance from B to AC) and in the second case they have the same height (the distance from D to AB and BC).

[@HARSHVARDHAN-im2hb]

Speed 1.25x

[@yourmajestythequeen8147]

Harsh Vardhan noob, use 1.75

[@fatematuzzohra9322]

Me 2 on test day for final review.

[@yeshwantghotkar2924]

Do for external bisector of an angle of a triangle plz

[@sashasanti4771]

yes pls

[@TheRameswaran]

nice 👍

[@paolasaenz1737]

Khan Academy can you please help me figure out the difference between a postulate and a definition's meaning in geometry please. I have test this week and I really need the help.

[@thomash7100]

rip

[@m.c.ravioli1521]

I think you need to intertwine the outer bisectors for the answer...

[@Its_my_plan]

sir i didn"t understand that part when you said AD = CD . The question is , is it necessary if you bisect an angle then the segment opposite to the angle will split into 2 equal part?

[@christinabaus5544]

nice

[@prafullohale5635]

khan sir you r a scholar

[@silverstep6251]

Thx man you just saved my day

[@namir9011]

Can we produce AB and join it with the line constructed by us from C which is parallel to BD

[@brokelama7191]

Thanks, it helped :D

[@drevil37373]

isnt it impossible to manipulate the length of CF. If you have that the angle of FDC is defined and the length of DC is defined and then if Angle DCF is at point C and is parallel to line BD, doesnt that mean that the angle has to already be defined so the only way the line is equal to BC, if the shape is a rhombus or did i miss something here?

[@tomdekler9280]

Only rhombuses have angle bisectors as diagonals so that makes sensem

[@markjackson6642]

Wow!!!

[@mirzabayev]

Thank you

[@Akash_R_Sharma]

thanks sal

[@anshbhatt2457]

Fellow man, I call from the future. I comment in 2016.

[@FirstNameLastName-fu8ml]

Ansh Bhatt lol... from the future 😂😂😂

[@lezzymarkus2924]

No you are from the past its 2017 now lol

[@williamsingh8335]

Ansh Bhatt i think you should first learn english

[@prajaktajadhao1211]

MScience 😂😂

[@rubikscubing5634]

@@lezzymarkus2924 no its 2018

[@soumisil8285]

If two straight lines are perpendicular to different straight lines respectively , then prove that the angle between the two staight lines are equal to that of the perpendiculars.can anyone help me prove this theorem?

[@tanishasagar9617]

Your english so good than that of maths .....Yes but maths is also nice as well

[@mirzabayev]

Good

[@Rastalogy]

How do you do it? 😢😢

[@AhmadKamalKhanOfficial]

How CF=BC?

[@jfusion99]

Notice that the triangle BCF (the bigger one) is isosceles because angles CFD and CBD are equal.

[@AhmadKamalKhanOfficial]

thanks

[@sashasanti4771]

youre welcome

[@debajyotisaha5523]

Correct pronunciation

[@TheSdawg0311]

First! Thank You!

[@skillhhy8707]

32th

[@rushilpatel7418]

*32nd

[@alseotte]

Sevens bith

[@rhcp4565]

Second! You Rock!

[@jonathanaugustin647]

I'm confused, isn't angle ADB = to angle CDB since it's a bisector?

[@tanishasagar9617]

Hey guys.... Try to listen him at 0.25x ☺☺☺too slow

[@AmanKumar-te8ft]

TriangleFDC is something which we call JUGAAD in India. Thanks BTW.

[@mohammadp_ahvazi4526]

We cant say that BC is equal to FC we proved that abc and fcd are similar

[@sashasanti4771]

what

[@ilie_trolebuz_ratbv]

So the Angel ADB=Angle CDB?

[@joninfect8681]

4:55 wee wee =P jokes *sorry

[@ravuthsim]

what kind of electronic writing pad do you use for this training?

[@gun6883]

you are wrong, refer book to prove (IN YOUR DIAGRAM) AB/BC = AD/DC

[@FPSGHoST808]

sixth! u ಠ_ಠ !

[@cezersotomayor3239]

what?

[@MrVHI123]

Forth! You Scissors! You Rock too!

[@Drone_Pro234]

Let go my hw is complete😂

[@vennilavennila6413]

English bad improve it if it was good video can be understand well

[@rabeehriyas6039]

Your grammar is very good😂😂😂

[@jackdagerman302]

Rabeeh Riyas r/wooosh

[@tanishasagar9617]

@@rabeehriyas6039 that's what I'm noticing 😇😂😂😎😎😎🤗

[@abbasbookwala]

People who live in glass houses don't throw stones at others...

[@codeobey6821]

30th

[@johncyprus9212]

So chat

[@ilikeboys2944]

69th 😏😏

[@sath6089]

I still dont know how it works....

[@verainsardana]

he explained it so slowly i almost slept

[@tarunajoshi3933]

You are a stupid/idiot

[@chooha]

@@tarunajoshi3933 and you are a typical indian with a fixed narrow mindset ? Are you ?

[@chooha]

You can learn it, you just have to : 1. Build your base right, get the "intuition" of the theorums that he talks about in this video 2. Get a pen and paper, draw the diagrams yourself, draw different triangles and draw a bisector to them, *play around* with it Remember, you can learn everything, you just need to try a few times and not give up, and it needn't be boring ! As Elon Musk puts it, when you are studying you are essentially downloading algorithms and data into your brain, and the process must be more fun than it is currently in our system. I know this reply might be late for you but it is for anyone who gets stuck and reads it.

[@tarunajoshi3933]

@@chooha 🖕

[@Abhijit-fw6jh]

Thank you sir

[@dekinora]

Thanks

[@57SwayamMukherjee]

Thank You Sir

[@noorbahaa7907]

Thank you so much