Nice solution. Simpler than mine, which went as follows: Let AD = x, CD = y. We need to find x. Using Law of Sines in △BCD, we get: sin C / 3 = sin 2θ / y = 2 sin θ cos θ / y sin C = 6 sin θ cos θ / y Using Law of Sines in △ABC, we get: sin C / 6 = sin θ / 4 sin C = 3 sin θ / 2 So we get: 6 sin θ cos θ / y = 3 sin θ / 2 cos θ = y/4 Using Law of Cosines in △ABD, we get: 3^2 = 6^2 + x² − 2(6)(x) cos θ 9 = 36 + x² − 12x(y/4) 0 = 27 + x² − 3xy 3xy = x² + 27 y = (x²+27)/(3x) Using Law of Cosines in △ABC, we get: 4^2 = 6^2 + (x+y)² − 2(6)(x+y) cos θ 16 = 36 + (x+y)² − 12(x+y) (y/4) (x+y)² + 20 − 3y(x+y) = 0 Substitute in y = (x²+27)/(3x) → x+y = (4x²+27)/(3x) ((4x²+27)/(3x))² + 20 − 3((x²+27)/(3x))((4x²+27)/(3x)) (4x²+27)²/(9x²) + 20 − 3(x²+27)(4x²+27)/(9x²) Multiply through by 9x² (x²+27)² + 180x² − 3(x²+27)(4x²+27) = 0 4x⁴ − 9x² − 1458 = 0 (x² + 18) (4x² − 81) Since x is a real positive number, then: x² = 81/4 *x = 9/2*
Triangle EBC similar to BAC. So BE=6y; Triangle BED similar to AEB so 2=BE:DE=AE:BE;AE=2*BE=12y; AC=12y+4y=16y In similar triangles EBC and BAC, 4:16y=4y:4 ; 64*y^2=16; y*y=1/4 ; y=1/2 AD=AE-DE=12y-3y=9y=9/2 The basic angle bisector theorem is used to get the ratio 3y and 4y and then nothing else is used after that except the similarity arguments.
Καλημέρα σας. Επειδή στα Ελληνικά βιβλία δεν αναφέρεται ο τύπος AD^2 = AB*AC-BD*CD, θα μπορούσατε να μου υποδείξετε μία απόδειξη σε αυτόν και αν είναι "επώνυμο" θεώρημα, το όνομά του; Ευχαριστώ.
I agree. We are owed an explanation about this unfamiliar equation. It was invoked in two RUclips videos [explaining the same problem] with no explanation. I searched the Internet with no success.
Uz BC pagarinājuma atliekkam punktu X tā , ka XAB=BAC=ʘ , tad XAB ~ CBD , AX/AC=3x/4x , XB=4 , AB - bisektrise AB=√(3x*4x-3*4), x=2, XA=6, AC=8, XAB~CBD k=2, DC=XC/2=3.5 AD=8-3.5=4.5!
تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا . تحياتنا لكم من غزة فلسطين .
Nice solution. Simpler than mine, which went as follows:
Let AD = x, CD = y. We need to find x.
Using Law of Sines in △BCD, we get:
sin C / 3 = sin 2θ / y = 2 sin θ cos θ / y
sin C = 6 sin θ cos θ / y
Using Law of Sines in △ABC, we get:
sin C / 6 = sin θ / 4
sin C = 3 sin θ / 2
So we get:
6 sin θ cos θ / y = 3 sin θ / 2
cos θ = y/4
Using Law of Cosines in △ABD, we get:
3^2 = 6^2 + x² − 2(6)(x) cos θ
9 = 36 + x² − 12x(y/4)
0 = 27 + x² − 3xy
3xy = x² + 27
y = (x²+27)/(3x)
Using Law of Cosines in △ABC, we get:
4^2 = 6^2 + (x+y)² − 2(6)(x+y) cos θ
16 = 36 + (x+y)² − 12(x+y) (y/4)
(x+y)² + 20 − 3y(x+y) = 0
Substitute in y = (x²+27)/(3x) → x+y = (4x²+27)/(3x)
((4x²+27)/(3x))² + 20 − 3((x²+27)/(3x))((4x²+27)/(3x))
(4x²+27)²/(9x²) + 20 − 3(x²+27)(4x²+27)/(9x²)
Multiply through by 9x²
(x²+27)² + 180x² − 3(x²+27)(4x²+27) = 0
4x⁴ − 9x² − 1458 = 0
(x² + 18) (4x² − 81)
Since x is a real positive number, then:
x² = 81/4
*x = 9/2*
Belíssima solução. Obrigado, mestre.
Triangle EBC similar to BAC. So BE=6y; Triangle BED similar to AEB so 2=BE:DE=AE:BE;AE=2*BE=12y; AC=12y+4y=16y
In similar triangles EBC and BAC, 4:16y=4y:4 ; 64*y^2=16; y*y=1/4 ; y=1/2
AD=AE-DE=12y-3y=9y=9/2
The basic angle bisector theorem is used to get the ratio 3y and 4y and then nothing else is used after that except the similarity arguments.
this is a sophisticated nested calculation!
10 l1=3:l2=4:l3=6:sw=.01:goto 110
20 dgu1=l3/l2*sin(w):dgu2=sin(pi-3*w-w2)
30 dg=dgu1-dgu2 :return
40 w2=sw:gosub 20
50 dg1=dg:w21=w2:w2=w2+sw:if w2>pi then stop
60 w22=w2:gosub 20:if dg1*dg>0 then 50
70 w2=(w21+w22)/2:gosub 20:if dg1*dg>0 then w21=w2 else w22=w2
80 if abs(dg)>1E-10 then 70
90 wu=pi-2*w-(pi-3*w-w2):dfu1=l2/l1*sin(pi-3*w-w2):dfu2=sin(wu)
100 df=dfu1-dfu2 :return
110 w=sw:gosub 40
120 df1=df:wu1=w:w=w+sw:if w>pi then stop
130 wu2=w:gosub 40:if df1*df>0 then 120
140 w=(wu1+wu2)/2:gosub 40:if df1*df>0 then wu1=w else wu2=w
150 if abs(df)>1E-10 then 140
160 print w:lx=l1*sin(w2)/sin(w):print"x="; lx:rem probe
170 l3r=lx*cos(w)+l1*cos(w2):print l3r:fe=(1-l3/l3r)*100:print "der fehler ist=";fe;"%"
180 dim x(2),y(2):mass=1500/(l1+l2+l3):goto 200
190 xb=x*mass:yb=y*mass:return
200 x(0)=-l3*cos(w2+2*w):y(0)=0:x(1)=x(0)+l2:y(1)=0:x(2)=0:y(2)=l3*sin(w2+2*w)
210 x=x(0):y=y(0):gosub 190:xba=xb:yba=yb:xbe=xb:ybe=yb:for a=1 to 3:ia=a:if ia=3 then ia=0
220 x=x(ia):y=y(ia):gosub 190:xbn=xb:ybn=yb:goto 240
230 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return
240 gosub 230:next a:dx=l1*cos(2*w):dy=l1*sin(2*w):xf=x(0)+dx:yf=y(0)+dy
250 xba=xbe:yba=ybe:x=xf:y=yf:gosub 190:xbn=xb:ybn=yb:gosub 230:x=x(2):y=y(2)
260 gosub 190:xbn=xb:ybn=yb:gcol 9:gosub 230
0.50536051
x=4.5
6
der fehler ist=-1.51406981E-8%
>
run in bbc basic sdl and hit ctrl tab to copy. the graphics shows the real angles
Thank you.Good explanation
Let angle(BDC) = lambda and draw a perpendicular from B to AC intersecting at E implies BE = 3 * sin(lambda) = 6 * sin(theta) implies cos^2(lambda) = 1 - sin^2((lambda) = 4 * cos^2(lambda) - 3
and 3 * sin(lambda) = 4 * sin(2 * theta + lambda) implies 3 = 2 * (2 * cos(theta) * sqrt(4 * cos^2(theta) - 3) + 4 * cos^2(theta) - 2) implies
(7 - 8 * cos^2(theta))^2 = 16 * cos^2(theta) * (4 * cos^2(theta) - 3) implies 49 = 64 * cos^2(theta) implies cos^2(theta) = 49/64 implies cos(theta) = 7/8 implies
9 = x^2 + 36 - 12 * x * (7/8) using law of cosines on triangle(ABD) implies 2 * x^2 - 21 * x + 54 = 0 implies (2 * x - 9) * (x - 6) = 0 implies x = 9/2, x = 6.
x = 6 implies Triangle(BAD) is isosceles triangle implies m(angle(ABD)) = m(angle(ADB)) = 180 - lambda implies lambda = 90 + theta/2 implies sin(90 + theta/2) = cos(theta/2) = 2 * sin(theta)
implies 1 - cos(theta) = 8 * sin^2(theta) implies 8 * cos^2(theta) - cos(theta) - 7 = 0 implies (8 * cos(theta) + 7) * (cos(theta) - 1) = 0 implies cos(theta) = 1 or cos(theta) = -7/8 implies theta = 0 or theta > 90 implies 2 * theta > 180, therefore we drop x = 6 and x = 9/2.
Nice and awesome, many thanks, Sir, this is great!
∆ABC → AB = 6; BC = 4; AD = 3; AC = AD + CD = x + CD; CAB = φ; DBC = 2φ →
EBC = DBE = φ; BE = b; x = ?
3/4 = DE/CE → DE = (3/4)CE → CE = 4z → DE = 3z → b^2 = 12 - 12z^2
∆ABC ≈ ∆BCE → 4/6 = 4z/b → 4b = 24z → b = 6z →
b^2 = 36z^2 = 12 - 12z^2 → z = 1/2 → 7z = 7/2 →
3/4 = 6/(x + 7/2) → x = 9/2
btw:
AC = x + 7z = 8
49/4 = 9 + 16 - 2(3)(4)cos(2φ) = 25 - 24cos(2φ) →
cos(2φ) = 17/32 → sin(2φ) = 7√15/32 → cos(φ) = 7/8 → sin(φ) = √15/8
ABC = ϑ → cos(ϑ) = -1/4
Καλημέρα σας. Επειδή στα Ελληνικά βιβλία δεν αναφέρεται ο τύπος AD^2 = AB*AC-BD*CD, θα μπορούσατε να μου υποδείξετε μία απόδειξη σε αυτόν και αν είναι "επώνυμο" θεώρημα, το όνομά του; Ευχαριστώ.
I agree. We are owed an explanation about this unfamiliar equation. It was invoked in two RUclips videos [explaining the same problem] with no explanation. I searched the Internet with no success.
Uz BC pagarinājuma atliekkam punktu X tā , ka XAB=BAC=ʘ , tad XAB ~ CBD , AX/AC=3x/4x , XB=4 , AB - bisektrise AB=√(3x*4x-3*4), x=2, XA=6, AC=8, XAB~CBD k=2, DC=XC/2=3.5 AD=8-3.5=4.5!
Great one!
Спасибо!
ด้านที่ให้หาเท่ากับสามคูณรูทสาเพราะด้านที่เหลือคือ หกและ สาม เป็นสามเหลี่ยมมุมฉากมีมุมซีต้าเท่ากับ30องศาครับ
ด้านตรงมุม badหารด้วยด้านตรงมุมadb เท่ากับ3หรด้วย6เท่ากับ1/2เท่ากับ
Sinซีต่้าเท่ากับ30ิองศา
สามเหลี่ยมabdเท่ากับเป็นสามเหลืยมมุมฉาก
มีมุมadbเป็นมุมฉากด้านadยาวเท่ากับ3คูณด้วยรากที่3 ครับ
Without pen: ruclips.net/video/hU7jERYj04s/видео.html
we have to prove that the triange ABD is similar to the triangle ABC.t hen AD/3 = 6/4. AD - 9/2
Nah they aren’t similar
why not?@@gummy8643
unfortunately, i could not prove this. (
@@ludmilaivanova1603 The point is you “find” two triangles similar but not that you “assume” two triangles similar.
that is why I said you have to prove this@@gummy8643