ANGLE BISECTOR! Find X Value by using the Angle Bisector Theorem & Law of Cosines | Simple Tutorial

Learn how to find the length of the angle bisector of a triangle using two different methods. Quick and easy explanation by PreMath.com

[@noahhysi8622]

Amazing video! It taught me the angle bisector theorem 👍

[@PreMath]

So nice of you Noah! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃

[@이경섭-m2b]

@@PreMath ㅇㅇ

[@xXDiver12Xx]

It's 1AM and i refuse to sleep without solving that one math problem. And this, my brothers, is exactly what i was searching.

[@عمرانآلعمران-و7خ]

Great video I can give a third solution to yours. Ii’s simply the area of the triangle ABC = the sum of areas of the inner triangle , then apply Heron formula to each triangle, we end up with a radical equation that can be solved for x.

[@philipkudrna5643]

It would be nice to show where the angle bisector theorem comes from. I definitely need to remember Al-Quashi‘s law of cosines, we didn‘t learn it in school, but I find it very useful! The Angle bisector theorem is, of course, quicker, but cannot be applied in all situations... Thank you for that neat little problem and the easy to follow explanations!

[@PreMath]

Thanks Philip for nice feedback. I'll try to make a video on Angle Bisector Theorem proof pretty soon! You are awesome 👍 Take care dear and stay blessed😃

[@Ramkabharosa]

Proof of Angle Bisector Theorem: By the Sine Rule, sin(ADC)/|AC| = sin (ACD)/|AD| (Eq.1) and sin(BDC)/|BC| = sin (BCD)/|BD| (Eq.2). Since sin(ADC) = sin(BDC) (supplementary angles) and sin (ACD) = sin (BCD) (equal angles), dividing Eq.2 by Eq.1 gives us |AC|/|BC| = |AD|/|BD|. So |AC|/|AD| = |BC|/|BD| (cross exchanging), i.e., a/b = c/d and we are done. Trigonometry is Empress when it comes to Geometry! . Proof of x² = ab - cd: By the Cosine Rule, c² = x² + a² - 2ax.cos(ACD) (Eq.1) and d² = x² + b² - 2bx.cos(BCD) (Eq.2). Since cos(ACD) = cos(BCD), b.(Eq.1) - a.(Eq.2) gives us b.c² - a.d² = (b-a).x² + b.a² - a.b² = (b-a).x² - (b-a).ab. And since ad = bc, we further get b.c² - a.d² = ad.c - bc.d = - (b-a).cd = (b-a).x² - (b-a).ab. Hence (b-a).ab - (b-a).cd = (b-a).x². Thus (b-a).x² = (b-a).(ab - cd) and so x² = ab - cd, provided (b-a) is not zero. But if b-a = 0 then a=b & c=d, so we get from Pythagoras' Theorem that x² = a² - c² = ab - cd. Algebra is Queen! .

[@YogeshSharma-ys3hm]

@@Ramkabharosa hey bro thanku for the explanation but can you solve this question I am getting square root as negative by the cosine method whole applied iff. AB=8 , AC=8 ,BC=12 AD is an angle bisector then what is AD?

[@Ramkabharosa]

@@YogeshSharma-ys3hm This is an almost trivial problem because if |AB| = |AC|, then AD will be the bisector of angle CAB and also the perpendicular to BC. So, |AD|² would be 8² - 6² = 64 - 36 = 28. Hence |AD| would be √28 = 2√7. Perhaps, you meant to say that CD is the angle bisector of ACB. Then the formula in the video would give you that a = |AC| = 8, b = |BC| = 12, & c = |AB| = 8. And we would get 2/3 = 8/12 = a/b = m/(8 - m). So 16 - 2m = 3m and thus 5m = 16. So m = 16/5 = c and 8 - m = 24/5 = d. Hence x² = ab - cd = 8(12) - 16(24)/25 = 96.(1 - 4/25) = 96(21)/25 = (16)(9)(14)/25. So x = (12/5)√14. .

[@liamdacre1818]

I prefer the first method. You explained it very well and it’s much clearer now

[@evanj3535]

I used the law of cosines to get Angle BCA and Angle CAB. Angle DCA is half of angle BCA, and 180 - Angle BCA - Angle CAB = Angle CDA. Then I used the law of sines to get CD.

[@PreMath]

Thanks Evan dear for the feedback. You are awesome 👍 Keep smiling😊

[@walker55able]

Again impressive i wasn't aware of method 1 formula which seemed less involved!

[@sastipadasadhu2254]

I have iearnt many types maths from your channel, thanks for you

[@aGuyWithConscience]

Would you prove x^2=ab-cd, please?

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@theodoresweger4948]

Thank you it has been a long time since I had a class in geometry...

[@ExpressStaveNotation]

So what is the theorem that "x^2 = ab - cd" called, and how do you prove this?

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@tolikbror_1927]

I study in Russia and i remember this theorem Thank you

[@최희준-f9r]

I visit this site from time to time. This site is very interesting to me. Because I can think of myself 40 years ago. But I don't know why X² is the same as ab-cd in this problem. I want to know the reason why X² is the same as ab-cd in this problem. Thanks for reading it.

[@HappyFamilyOnline]

Great video👍 Thank you so much 😀

[@PreMath]

You are so welcome! Cheers😀

[@someoneistyping]

Please help me out to prove this equation, x² = ab - cd?

[@devsgkquizdevakrishnan9308]

Very interesting.Can you please say which app you use to make videos?

[@PreMath]

So nice of you Dev dear! You are awesome 👍 I'm glad you liked it! We use Camtasia TechSmith utility! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃

[@leonardogu4305]

Thanks I’m learning this right now and that formula is not in our text book so I’m not sure I will use it. Thanks though

[@garypaulson5202]

Fun video sir, thank you

[@ebi2ch]

Construct a rectangle with AB as an edge and C as a path through it, and let E and F be the vertices of the edge through C, respectively. At this point If CF=a and BF=b, then (12-a)^2+b^2=100 and a^2+b^2=64. If we solve this simultaneous equation, we get a=9/2, and b^2=175/4. If we draw a vertical line from C to AB and set the intersection point as G, we get DG=(16/3)-(9/2)=(5/6). So x^2=(5/6)^2+b^2=400/9. x=20/3.

[@PreMath]

Great tip dear! You are awesome 👍 Take care dear and stay blessed😃

[@diablo888]

still needed to calculate 16/3 first

[@asadmuyinda3111]

After finding M , use Stewart’s theorem to get X . Thanks for the explanation

[@holyshit922]

I used law of sines and law of cosines Do we have isosceles triangle here

[@tomcruise6738]

I knew the second method but didn't know the first one. Alternatively I knew the direct formula to find the length of angle bisector and that is, {Root of 2(10*8*15*3)}/(10+8)=20/3 Where 15 is the semi perimeter and 3 came by semi perimeter 15 minus 12, the side on which the angle bisector lies.

[@auridannr]

Please. What is the origin of the formula x² = ab - ac?

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@kaliprasadguru1921]

Sir, will you be kind enough to give the proof of X² = a.b - c.d.

[@bwahf4685]

Take à look at the section 'proof 2' to find out the solution ⟹ proofwiki.org/wiki/Length_of_Angle_Bisector 👍

[@kaliprasadguru1921]

Thank you sir . No doubt it gives the length of angle bisector . But it is not the proof of the formula used in step 2 in this video ie. X² = a.b- cd. Regards .

[@bwahf4685]

@@kaliprasadguru1921 Notice that the demonstrated step ' d² = bc-BD.DC' is present in the 'proof 2' demonstration, don't pay attention to the final result... just this step that is proved. 😉

[@kaliprasadguru1921]

Got it . Many many thanks .

[@mryip06]

you can use the following 2 points to prove that. 1. cos law twice with the 2 angles (let's use θ and 180°-θ to denote them) on the base with length of 12. 2. a/b = c/d

[@marhsfirst]

Nice work

[@ChocoBunnie]

I know this is an old video but thank you! i needed this formula for a garment im sewing together and I couldnt for the life of me remember how to do it 😂

[@Aryan_Giri01]

3:47 what is the proof of x² = ab-cd ?

[@bwahf4685]

Take à look at the section 'proof 2' to find out the solution ⟹ proofwiki.org/wiki/Length_of_Angle_Bisector 👍

[@Aryan_Giri01]

@@bwahf4685 Thanks bro 🙏🙏🙏

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@ashokme1976]

Could you share the proof for the theorem: x^2=(a*b)-(c*d) ?

[@bwahf4685]

Take à look at the section 'proof 2' to find out the solution ⟹ proofwiki.org/wiki/Length_of_Angle_Bisector 👍

[@ashokme1976]

Thank you

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@ROCCOANDROXY]

Using the angle bisector theorem a/b = c/d implies ad - bc = 0. Your actually using the law of cosines to derive x^2 = ab - cd. Let m(angle(ACD)) = m (angle(DCB)) = theta. Using law of cosines on triangle(ACD) and triangle(DCB) implies c^2 = a^2 + x^2 - 2axcos(theta) d^2 = b^2 + x^2 - 2bxcos(theta) implies b|(2axcos(theta) = a^2 + x^2 - c^2) -a|(2bxcos(theta) = b^2 + x^2 - d^2 implies (a - b)x^2 = a^2b - ab^2 + d^2a - c^2b = ab(a - b) + ad^2 - dbc + acd - bc^2 + dbc - acd = ab(a - b) + d(ad - bc) + c(ad - bc) - dc(a - b) = (a - b)(ab - cd) implies x^2 = ab - cd. Deriving the angle bisector theorem: Let Let m(angle(CDA)) = lambda implies m(angle(CDB)) = 180 - lambda. Area(triangle(ACD))/Area( triangle(DCB)) = 1/2 * a * x * sin(theta)/(1/2 * b * x * sin(theta)) = a/b = 1/2 * c * x * sin(lambda)/(1/2 * d * x * sin(180 - lambda)) = c/d implies a/b = c/d. In general, letting AC = a, CB = b and AB = c with AD = y implies DB = c - y and the angle bisector CD = x and m(angle(ACD)) = m (angle(DCB)) = theta. a/b = y/(c - y) implies y = ac/(a + b) implies c - y = bc/(a + b) implies y(c - y) = abc^2/(a + b)^2 implies x^2 = ab((a + b)^2 - c^2)/(a + b)^2 implies x = sqrt(ab((a + b)^2 - c^2)/(a + b)^2).

[@rangaswamyks8287]

X^2=ab-cd Could u please give proof sir Yours.... Swamy Thank u sir.. You solved it beautifully and easilly

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@c.mohanchandrasekaran8166]

call the half angle of bisected angle to be y. use the idea of area of a triangle is 1/2xbcxSinA the two t triangles, and also for the whole triangle, Find the area of triangle using the formula square root of [ d(s-a)(s-b)(s-c)] , s =a+b+c/2, (P) Equate them , thereby you will get SinA = 3X SQUARE ROOT OF7/8. Using this you can calculate SinA/2 = Square root of 7/4.(1) The sum of the areas of the two triangles is 9Xx= 5 x square root of7/3 SinA/2( this is obtained by equating this value with the area got using(P) [2], substituting (1) in [2], we get the value X = 20/3. May be this method is laborious.

[@johnnath4137]

There is a formula for the angle bisector: AD² = bc(1 - a²/(b + c)²) = )10 x 8)(1 - (12²/(10 + 8)²) = 80(1 - 80/324) =80 x 244/324 = 80 x 5/9 = 400/9 ⇒ AD = 20/3.

[@kalyanbasak6494]

Namaskar sir,x=8.23333unit I have tried sir thanks u r genius

[@ASHAIKH1]

Wrong because you said AD is congruent to BD by using sign of congruency, so they have to be equal.

[@johnbrennan3372]

Two applications of the cosine rule would be my preferred method. First method presupposes knowledge of formulae which I was not aware of, but would love to know how they are derived.

[@PreMath]

Thanks John for the visit! You are awesome 👍 Take care dear and stay blessed😃 Kind regard

[@jamesdo3841]

The very first thing to determine is AD and BD before you can proceed with the 2 methods. In your video, you determined AD and BD under the angle bi-sector method.

[@devondevon4366]

Answer = 6.6666 or 20/3 according to the angle bisector theorem, the line of length 12 will have the same ratio as 10 and 8 and those two numbers 20/3 (6.666) and 16/3 (5.333) since 20/3 over 16/3=20/36 or 10/18. That is, 6.666/5.333 = 10/8 or 1.25 Using SSS (10, 8, and 12) the angle that bisected = 82.819 degree which implies half = 41.4095 degree Using SAS and the sides 10, 6.666 and angle 41.4095 yields 6.6666

[@PreMath]

Awesome my friend😀

[@soufianeaitabbou3727]

I found an other way to find x is to calculate the cos of the angle (ACD) in the triangle ACD on the function of x; then calculate the cos of the angle (DCB) in the triangle ABD on the function of x; and we know that the two angles are equals (because of the bisector) so we are going to find an equation where the unknown is x ; and we are going to find the same result.

[@ΑλέξανδροςΓιακαλής]

Heron formula to find the area of the triangle using the semi perimeter. Then knowing the base and the area of the triangle you can find the height of the triangle. Then use Pethagorem on one of the 2 triangles and then cosin law

[@HafsaSikanderKhan]

ماشاءاللّٰه ❤

[@PreMath]

Thanks dear!

[@gemalbenallie1007]

I watched and liked the video

[@PHANTOMFACTS10]

how is x2 = ab-cd

[@242math]

both work but I prefer using the angle bisector theorem

[@PreMath]

Thanks my dear friend for your candid feedback. You are awesome 👍 Take care dear and stay blessed😃

[@kamarinelson]

I use the law of cosines to find the angle opposite the side of length 12 symbolically. Then I calculated the areas of all 3 triangles symbolically as well. Knowing that the areas of the 2 smaller triangles adds to the area of the larger one, that gives us the 1 equation needed to solve for x. I didn't solve for anything other than the angle and x.

[@kamarinelson]

I had to make use of the double angle formula, which further demonstrated why I needed to solve for the angle specifically.

[@dhrubajyotidaityari9240]

∆ABC, a/b=10/8, a+b=12, a=20/3 CosB=(10²+400/9-x²/(2.10.20/3) from ABD CosB =10²+12²-8²)/(2.10.12), from∆ABC. Equating, x=20/3

[@raulcastrosanchez5322]

aplico teorema de la bisectriz y stewart y sale en dos patadas

[@shashwatvats7786]

This question is only of two steps 1. Application of angle bisector theorem. 2.construction of altitude from c on ab and applying the pythagoras theorem and then one line simplification.

[@kevinmadden1645]

X also is equal to 25/3 . Use Law of Cosines on each triangle.. Ambiguous Case of the Law of Sines ..

[@HafsaSikanderKhan]

لَا إِلٰهَ إِلَّا اللّٰهُ مُحَمَّدٌ رَسُولُ اللّٰهِ‎ ❤

[@PreMath]

So nice of you Hafsa dear! You are awesome 👍 I'm glad you liked it! Please keep sharing premath channel with your family and friends. Take care dear and stay blessed😃

[@arturovinassalazar]

But these formula used at fist method isnt Stewart Theorem!!

[@billrundell2097]

In the drawing it shows AD is = DB AD = 6 and db=6

[@PreMath]

Dear Bill, CD is an angle bisector, not a median! Therefore, we can't say AD=BD Thanks dear for you input. You are awesome 👍 Take care dear and stay blessed😃

[@billrundell2097]

@@PreMath In the drawing you labeled ad= db You used the double slash lines on both ad=db But as you said, you inferred it to be an angle bisector. Your procedure of angle bisector is correct.

[@sakshamsingh1778]

I used stewart + angle bisector theorum

[@poppyaustin7315]

Done, but with your hints

[@jaaaayt.20]

watched and liked the video

[@asadmuyinda3111]

Immediately after Bisector theorem ; use Stewart’s theorem

[@shrikantaroy6711]

Require to show ,x2=ab-cd please.

[@triathlon.75]

kzbin.info/www/bejne/ZovaaHiKeNyVpqc

[@WaiWai-qv4wv]

Thanks

[@PreMath]

Welcome dear Thanks dear for the feedback. You are awesome 👍 Take care dear and stay blessed😃

[@WaiWai-qv4wv]

Very thanks. How are you?

[@rangaswamyks8287]

You saying fill in the blanks No... You have to say "let us substitute the values" sir

[@renatosouza2343]

Very well.

[@tgx3529]

((1/2) sin alfa*x*8)+((1/2)* sin alfa*x*10=(1/2)* sin(2 alfa)*10*8. We have then (cos alfa)*160/18=x. From cosine theory se have cos(2alfa)=1/8====> cos alfa=sqrt((1+1/8)/2)

[@luigipirandello5919]

Great

[@leumasanogyygona5400]

Nice.

[@charlesbromberick4247]

8-10-12 is NOT a right triangle; 6-8-10 is.

[@ASHAIKH1]

You used sign of congruency in wrong place otherwise you're right.

[@jonthebob2159]

Or take sqrt((10x8)-(20/3 x 16/3)) and gets you 20/3

[@theophonchana5025]

x = Square root of (400÷9) = 20÷3

[@theophonchana5025]

#cosine #Trigonometry

[@rangaswamyks8287]

Using law of cosines is easy

[@devondevon4366]

x=6.6666 or 6 and 2/3

[@PreMath]

Thanks my dear friend for the feedback. You are right on! 👍 Take care dear and stay blessed😃

[@TechToppers]

Stewart Theorem + Angle Bisector Kill

[@فراسمعابره-ج5خ]

اذا كان cB=16/3 فإن الامر لا يحتاج إلى كل هذا العناء والتعب

[@theophonchana5025]

cos (angle) = 18÷24 = 9÷12 = 3÷4 = 0.75

[@sh555222]

رائع

[@rajendrasheregar3113]

Shift x + axis LevO sidE to righT angLE ----- when righT angLE is formeD at PoinT 5 of 12 besides midPoinT --- 8 is eveNumber noT oDD so righT angLE forms odd because odd + odd forms righT angLE constancY --- to eveNumber helDs righT angLE ----- thus x canT be 6 oR 7 becausE 8 doesnT change to 9 is x forMs 6 ---- thus aFteR shifted to righT angLE -- x remains 5

[@ЮрийЯкубовский]

есть готовая формула CD={√AC*CB(AC+CB+AB)*(AC+CB-AB)}/(AC+CB)=20/3

[@shreyanshpatel0707]

stewart's theorm

[@Ramkabharosa]

Proof of Angle Bisector Theorem: By the Sine Rule, sin(ADC)/|AC| = sin (ACD)/|AD| (Eq.1) and sin(BDC)/|BC| = sin (BCD)/|BD| (Eq.2). Since sin(ADC) = sin(BDC) (supplementary angles) and sin (ACD) = sin (BCD) (equal angles), dividing Eq.2 by Eq.1 gives us |AC|/|BC| = |AD|/|BD|. So |AC|/|AD| = |BC|/|BD| (cross exchanging), i.e., a/b = c/d and we are done. Trigonometry is Empress when it comes to Geometry! . Proof of x² = ab - cd: By the Cosine Rule, c² = x² + a² - 2ax.cos(ACD) (Eq.1) and d² = x² + b² - 2ax.cos(BCD) (Eq.2). Since cos(ACD) = cos(BCD), b.(Eq.1) - a.(Eq.2) gives us b.c² - a.d² = (b-a).x² + b.a² - a.b² = (b-a).x² - (b-a).ab. And since ad = bc, we further get b.c² - a.d² = ad.c - bc.d = - (b-a).cd = (b-a).x² - (b-a).ab. Hence (b-a).ab - (b-a).cd = (b-a).x². Thus (b-a).x² = (b-a).(ab - cd) and so x² = ab - cd, provided (b-a) is not zero. But if b-a = 0 then a=b & c=d, so we get from Pythagoras' Theorem that x² = a² - c² = ab - cd. Algebra is Queen! .

[@KudosJkp]

That's really great.

[@ionelpatriche6866]

Frumos!

[@reforma715]

👍👍👍

[@theophonchana5025]

x^(2) = 400÷9

[@anushkabhandari9508]

5.3333

[@davidbrisbane7206]

I found x. It is in the middle of the diagram 🤣😂.

[@theophonchana5025]

#lawofcosines

[@theophonchana5025]

x variable

[@theophonchana5025]

m = 20÷3

[@marciec6862]

👍🏻

[@anushkabhandari9508]

0.75

[@anushkabhandari9508]

44.4444

[@anushkabhandari9508]

180

[@anushkabhandari9508]

144

[@anushkabhandari9508]

4

[@giuseppemalaguti435]

8

[@anushkabhandari9508]

100

[@anushkabhandari9508]

6.6666

[@anushkabhandari9508]

1

[@rachajhie1097]

are you crazy?

[@leonardogu4305]

Thanks I’m learning this right now and that formula is not in our text book so I’m not sure I will use it. Thanks though

[@theophonchana5025]

x = Square root of (400÷9) = 20÷3

[@anushkabhandari9508]

0.75

[@anushkabhandari9508]

4

[@anushkabhandari9508]

100