In the acute triangle ABC, AB equals 14, BC equals 14, the area of triangle ABC is 84, find the value of Tanc and sinc I have worked it out, but the result is too complicated. If you want to verify it, please give the result

In the acute triangle ABC, AB equals 14, BC equals 14, the area of triangle ABC is 84, find the value of Tanc and sinc I have worked it out, but the result is too complicated. If you want to verify it, please give the result

BC*h/2=84, 14h=84, h=6.AB²-x²=h², x²=AB²-h²=196-36=160, x=4√10. AC²=h²+(14-4√10)²=36+196-112√10+160=392-112√10, AC≈6.15tanC=h/(14-x)=6/(14-4...

AB = AC, BD bisector angle B in acute isosceles triangle ABC: BC = AD + BD

The title is wrong and cannot be proved

It is known that ad, be are the heights of acute triangle ABC, a ` d ', B ` e' are the heights of acute triangle a ` B ` C ',
AB / ad = a ` B '/ a ` d', angle c = angle c '`=

From AB / ad = a'B '/ a'd' (similar to HL), we can get ∠ B = ∠ B '
Because ∠ C = ∠ C '
So △ ABC ∽ a'b'c '
So the corresponding height is proportional, that is, be / b'e '= the ratio of any corresponding edge (AD / a'd')
We can get ad · b'e '= a'd' · be
I'll have math problems in the future

As shown in the figure, △ ABC, ∠ ABC = 2 ∠ C, be bisection ∠ ABC intersects AC with E, ad ⊥ be with D, proving: (1) ac-be = AE; (2) AC = 2bd

It is proved that: (1) be bisection ∠ ABC, ∵ CBE = 12 ∠ ABC, ∵ ABC = 2 ∠ C, ∵ EBC = ∠ C, ∵ be = CE, ∵ ac-be = ac-ce = AE; (2) extend BD to N, make DN = BD, connect an. ∵ ad ⊥ be, ∵ ad bisection BN vertically, ∵ AB = an, ∵ n = ∵ ABN = ∵ NBC = ∥ C, ∵ an ∥ BC, ∥ C = ∥ NAC, ∥ NAC = ∥ n, ∥ AE = en, ∥ be = EC, ∥ AC = BN = 2bd

As shown in the figure, it is known that in the acute angle △ ABC, ∠ ABC = 2 ∠ C, the bisector of ∠ ABC and AD are perpendicular to D, and the verification is: AC = 2bd

It is proved that DF = De, AF, ∵ DF = De, ad ⊥ BF, ad = ad, ≌ △ ADF ≌ △ ade, (3 points) ≁ AE = AF, ∠ AFD = ∠ AEF ≌ ABC = 2 ∠ C, be is the bisector of ≁ ABC, ≁ Abe = ∠ C = ∠ EBC, ≁ AFE = {Abe +} BAF,} AEF = {EBC +} C} Fab =} ABF

In the acute triangle ABC: angle B = 2, angle c, be is the bisector of angle, ad is perpendicular to be and D, AC = 2bd is proved

Extend ad to cross BC with F, cross AC parallel line through D to cross BC with G,
Easy to know: ad = DF
=>DG is the median line of AFC, DG = 0.5ac
And angle DBC = 0.5, angle B = angle c = angle DGB
=> BD = DG
=> AC = 2BD

In the acute triangle ABC, known AB = 10, BC = 8, its area s = 32, find the length of the edge AC, thank you, God help

S = 1 / 2xabxbcxsinb = 32 SINB = 0.8 acute triangle, CoSb = 0.6 CoSb = (BC ^ 2 + AB ^ 2-ac ^ 2) / 2xbcx AB = 0.6 AC = 2 radical 17

As shown in the figure, in the acute angle △ ABC, BD ⊥ AC, de ⊥ BC, ab = 14, ad = 4, be: EC = 5:1, then CD=______ ．

∵BD⊥AC，∴∠ADB=∠BDC=∠BDE+∠CDE=90°，∵AB=14，AD=4，∴BD=142−42=65．∵DE⊥BC，∴∠BED=∠CED=90°，∴∠C+∠CDE=90°，∴∠C=∠BDE，∴△DEB∽△CED．∴DE：CE=BE：DE，CD：BD=CE：DE，∵BE：EC=5：1，∴...

The interior angles a and B of the triangle ABC are acute angles, and CD is high. If dB / ad = BC / AC, then the triangle ABC is high
A right triangle
B isosceles triangle
C isosceles right triangle
D isosceles triangle or right triangle
Non-B

Let AD / DB = (AC / BC) ^ 2 = K
(AC/BC)^2=(CD^2+AD^2)/(CD^2+BD^2)
=(CD^2+k^2AD^2)/(CD^2+BD^2)=k
k*CD^2+k*BD^2=CD^2+k^2BD^2
(k-1)CD^2=k(k-1)BD^2
That is, AC = BC
Or CD ^ 2 = KBD ^ 2 = ad * BD, that is CD: ad = BD: CD ⊥ CD ⊥ ab
The Δ ADC is similar to the Δ CDB
∠A= ∠DCB ∠B= ∠ACD
So the answer is "isosceles triangle or right triangle"

It is known that ad is the bisector of the interior angle of triangle ABC. Prove AC / AB = CD / DB

Make be parallel to AC and ad extension line at e through B,
Triangle QDC is similar to EBD,
AC/BE=CD/DB,
Ad is the bisector of the interior angle of the triangle ABC,
Angle bea = angle CAE = angle BAE,
AB=BE,
AC/AB=CD/DB.