robert@brl-tgr.ARPA (Robert Shnidman ) (11/07/85)
Given: Triangle ABC with angle bisectors AE and BD such that
AE=BD.
Prove: AC=BC.
C
/\
/ \
/ \
D/. .\E
/ .. \
/ . . \
/ . . \
/ . . \
/. .\
A/ ________________ \B
This problem is harder that it looks.yeff@Navajo.ARPA (11/09/85)
>Given: Triangle ABC with angle bisectors AE and BD such that >AE=BD. >Prove: AC=BC. > C > /\ > / \ > / \ > D/. .\E > / .. \ > / . . \ > / . . \ > / . . \ > /. .\ > A/ ________________ \B >This problem is harder that it looks. ok....here we go!! let AC=a,BC=B,AB=c,AD=BE=d now we all know that an angle bisector divides the side it hits into a ratio that is equal to the ratio of the sides.... so we have: AD/DC = AB/AC , BE/DE = BA/BC or, in terms of the lengths: d/(a-d) = c/b , e/(b-d) = c/a solve both eqns for c and we get: c = (bd)/(a-d) , c = (ad)/(b-d) set these equal, eliminate the d in each numerator, cross-multiply (feels so good!!), and we get: b^2 - bd = a^2 - ad take everything over to one side and factor and we get: (b - a) * (b + a - d) = 0 sooooo, either b - a = 0, or b+a-d=0... the second is impossible, bcuz a>d and b>0 (this is "intuitively obvious") so, this means that a - b = 0 ==> a = b ==> AC = BC qed aaaaahhhhhh!!!!! jeff hasn't used geometry in like 4 years..... it's nov 9, and it's 12:30 am.......do you know where your children are??
robert@brl-tgr.ARPA (Robert Shnidman ) (11/12/85)
> >Given: Triangle ABC with angle bisectors AE and BD such that > >AE=BD. > > >Prove: AC=BC. > > > > > C > > /\ > > / \ > > / \ > > D/. .\E > > / .. \ > > / . . \ > > / . . \ > > / . . \ > > /. .\ > > A/ ________________ \B > > > >This problem is harder that it looks. > > ok....here we go!! > > let AC=a,BC=B,AB=c,AD=BE=d > > now we all know that an angle bisector divides the side it hits into a ratio > that is equal to the ratio of the sides.... > > so we have: > AD/DC = AB/AC , BE/DE = BA/BC > > or, in terms of the lengths: > d/(a-d) = c/b , e/(b-d) = c/a > > solve both eqns for c and we get: > c = (bd)/(a-d) , c = (ad)/(b-d) > set these equal, eliminate the d in each numerator, cross-multiply (feels so > good!!), and we get: > b^2 - bd = a^2 - ad > take everything over to one side and factor and we get: > (b - a) * (b + a - d) = 0 > > sooooo, either b - a = 0, or b+a-d=0... > > the second is impossible, bcuz a>d and b>0 (this is "intuitively obvious") > > so, this means that a - b = 0 ==> a = b ==> AC = BC qed > > aaaaahhhhhh!!!!! > > jeff hasn't used geometry in like 4 years..... > > it's nov 9, and it's 12:30 am.......do you know where your children are?? AD=BE was NOT given! Solve the given problem.
gupta@asgb.UUCP (Yogesh K Gupta) (11/13/85)
> > take everything over to one side and factor and we get: > (b - a) * (b + a - d) = 0 > > sooooo, either b - a = 0, or b+a-d=0... > > the second is impossible, bcuz a>d and b>0 (this is "intuitively obvious") ^^^^^^^^^^ This is not necessarily true (a is less than d if a < c), but a+b-d=0 is still impossible as follows: a+b-d = AB + BC - AE = AB + BE - AE + EC Now, as ABE form a triangle, AB + BE > AE. . . . AB + BE - AE + AC > 0 -- Yogesh Gupta Advanced Systems Group, {sdcrdcf, sdcsvax}!bmcg!asgb!gupta Burroughs Corp., Boulder, CO. -------------------------------------------------------------------- All opinions contained in this message are my own and do not reflect those of my employer or the plant on my desk.
bill@ur-cvsvax.UUCP (Bill Vaughn) (11/14/85)
> >Given: Triangle ABC with angle bisectors AE and BD such that > >AE=BD. > >Prove: AC=BC. > > C > > /\ > > / \ > > / \ > > D/. .\E > > / .. \ > > / . . \ > > / . . \ > > / . . \ > > /. .\ > > A/ ________________ \B > > > >This problem is harder that it looks. > > ok....here we go!! > > let AC=a,BC=B,AB=c,AD=BE=d > > now we all know that an angle bisector divides the side it hits into a ratio > that is equal to the ratio of the sides.... > > so we have: > AD/DC = AB/AC , BE/DE = BA/BC > > or, in terms of the lengths: > d/(a-d) = c/b , e/(b-d) = c/a > (etc.) Sorry, but no cigar. The conditions of the theorem are AE = BD NOT AD=BE. One essentially has to PROVE that AD = BE because from this it follows easily that ABC is isosceles. (HINT: Proving the contrapositive looks like it should be easier. Thats the angle ( :-) ) I'm taking.) Bill Vaughn U of Rochester ur-cvsvax!bill@rochester.arpa
bs@faron.UUCP (Robert D. Silverman) (11/14/85)
> > > > take everything over to one side and factor and we get: > > (b - a) * (b + a - d) = 0 > > > > sooooo, either b - a = 0, or b+a-d=0... > > > > the second is impossible, bcuz a>d and b>0 (this is "intuitively obvious") > ^^^^^^^^^^ > This is not necessarily true (a is less than d if a < c), but > a+b-d=0 is still impossible as follows: > a+b-d = AB + BC - AE > = AB + BE - AE + EC > Now, as ABE form a triangle, AB + BE > AE. > . > . . AB + BE - AE + AC > 0 > > -- > Yogesh Gupta Advanced Systems Group, > {sdcrdcf, sdcsvax}!bmcg!asgb!gupta Burroughs Corp., Boulder, CO. > -------------------------------------------------------------------- > All opinions contained in this message are my own and do not > reflect those of my employer or the plant on my desk. It is my opinion that the proof is still incomplete. His original assertion that the angle bisector divides the side into the same ratio as the adjoining sides can hardly be called an axiom and is far from obvious. I, in fact hadn't remembered it as a theorem until the proof reminded me. To qualify as a rigorous proof either a reference should be made for this assertion or it should be proved separately as a lemma. I know of two other proofs. One is purely trigonometric and the other uses a theorem of Fermat (again far from obvious) that the 3 pairs of opposite sides of a hexagon which is inscribed in an ellipse , if extended, intersect at three points which are co-linear. The relation to an ellipse becomes obvious if you study the problem for a little bit. That's all I'm going to say. Bob Silverman (they call me Mr. 9)
leeper@mtgzz.UUCP (m.r.leeper) (11/16/85)
The distance from the line Px+Qy+R = 0 to the point (U,V) is
|PU+QV+R|/(P^2+Q^2)^(1/2)
This can be proved separately, and the interpretation is fairly
interesting, but let us apply it here by placing a Cartesian axis over
the problem with A at (0,0) and B at (0,1). Let the base angle at A be
2a and the base angle at B be 2b. So the line BC has the equation
y + (tan(2b))x - tan(2b) = 0
The line AC has equation
y - (tan(2a))x = 0
using the above formula, the length of segment AE is
tan(2b)/((1+(tan(2b))^2)^(1/2))
the length of segment BD is
tan(2a)/((1+(tan(2a))^2)^(1/2))
A little simple (careful not to lose roots) manipulation tells you that
2a = 2b. Hence the base angles of the triangle are equal so the
triangle is isoceles.
Mark Leeper
...ihnp4!mtgzz!leeperags@pucc-h (Dave Seaman) (11/17/85)
In a triangle with sides a, b and c the length of the angle bisector which
meets side c is given by
2
t = --- sqrt(abs(s-c))
c a+b
where s = (a + b + c) / 2. Note that "abs(s-c)" means "a*b*s*(s-c)" and
not the absolute value of (s-c).
I will post a proof of this formula as a separate lemma. By symmetry,
the corresponding formula for the bisector that meets side b is
2
t = --- sqrt(acs(s-b))
b a+c
Given t = t , we are to show that b = c.
c b
After clearing fractions we have
(a+c) sqrt (abs(s-c)) = (a+b) sqrt(acs(s-b)).
Squaring both sides and using the identities s-c = (a+b-c)/2 and
s-b = (a+c-b)/2:
2 2
(a+c) abs(s-c) = (a+b) acs(s-b)
or
2 2 2 2
(a +2ac+c )b(a+b-c) = (a +2ab+b )c(a+c-b)
which can be rearranged to
3 2 2 2 3 3 2 2 2 3
a b+a b +3ab c+b c = a c+a c +3abc +bc
or
3 2 2 2 2 2
a (b-c) + a (b -c )+3abc(b-c)+bc(b -c ) = 0
which factors as
3 2
(b-c)[a +a (b+c)+3abc+bc(b+c)] = 0.
Since a, b and c are all positive, the expression in brackets cannot be
zero. Therefore it is possible to divide, yielding b-c=0, or b=c.
--
Dave Seaman ..!pur-ee!pucc-h!agsven@hou2f.UUCP (V.VENKATESWARAN) (11/18/85)
Let angle A = 2a , angle B = 2b
By the sine rule on AEB get AB = AE sin(180-2b-a)/sin(2b)
By the sine rule on ADB get BD = AB sin(2a)/sin(180-2a-b)
BD = AE . sin2a . sin(2b+a)
------- ---------
sin2b sin(2a+b)
We will show that the only solution for sin2b sin(2a+b) = 1 is a = b.
----------------
sin2a sin(2b+a)
wlog let a < 45 , a <= b and let b = a+d
let f(d) = sin(2a+2d) sin(3a+d)
----------------------
sin2a sin(3a+2d)
f(0) = 1 .
if (2a+2d) <= 90 (2a+2d) > 2a ; else, 180-2a-2d > 2a
hence sin(2a+2d)
--------- > 1
sin2a
if 3a+d > 90 then sin(3a+d)
---------- also > 1
sin(3a+2d)
So examine f(d) for d in 0 to D = 90-3a
df/dd = [sin(3a+2d)[2cos(2a+2d)sin(3a+d) + sin(2a+2d)cos(3a+d)]
- 2cos(3a+2d)sin(2a+2d)sin(3a+d) ] / C
where C = sin(2a)sin(3a+2d)sin(3a+2d) > 0 for d in [0,D]
case1 2a+2d > 90
-----
Consider 2sin(3a+2d)cos(2a+2d)sin(3a+d) and
-2sin(2a+2d)cos(3a+2d)sin(3a+d)
-cos(3a+2d) > 0 and > cos(2a+2d)
sin(2a+2d) > sin(3a+2d)
The term sin(3a+2d)sin(2a+2d)cos(3a+d) >= 0 for d in [0,D]
So C.df/dd > 0
case 2 2a+2d =< 90
------
C.df/dd = sin(2a+2d)[sin(3a+2d)cos(3a+d)-cos(3a+2d)sin(3a+d)]
+sin(3a+d)[sin(3a+2d)cos(2a+2d)-cos(3a+2d)sin(2a+2d)]
+sin(3a+2d)sin(3a+d)cos(2a+2d)
= sin(2a+2d)sin(d) + sin(3a+d)sin(a) + sin(3a+2d)sin(3a+d)cos(2a+2d)
all quantities are >= 0 and sin(3a+d)sin(a) > 0
So for d in [0,D] df/dd > 0 . By the mean value thm. f(d) cannot equal
1 in (0,D]. So there is only one soln.
(Case 1 verifies that if angle B is obtuse, AE can never equal BD.)