In isoceles triangle ABC angle A=20, B=80, C=80.

Additional material

[ Editor: further solutions from rec.puzzles, May 1996, when the
problem went round as "triangle puzzle" and "Another IQ test problem"
(said to have "at least 6 solutions", to be done by geometric
construction without trigonometry). ]

In article <4maaqu$h8s@mugca.cc.monash.edu.au>,
tim@mugca.cc.monash.edu.au says...
>From one of the many IQ tests on the net:
>
>Take an isosceles triangle ABC.  Draw a line from C to a point on AB,
>call it D, so that BCD is 70 degrees and ACD is 10 degrees.  Draw a
>line from B to a point on AC, call it E, so that CBE is 60 degrees
>and ABE is 20 degrees.  What is the angle CDE ?

Let F be the intersection of AB with the line parallel to BC through E.
Let G be on the line through C and F with CG = CA (lengths equal) and
ACG = 20 degrees.  Since the line through C and D perpendicularly
bisects AG, AD = GD.  Since GAD = 60 degrees, triangle GAD is
equilateral - so, AD = AG.  Since AG=BC, AD=BC.
  Since EAB = ABE = 20 degrees, AE = BE.  Let H be the intersection of
BE and CF.  Then since triangles BCH and EFH are equilateral,
BE = BC + EF.  Thus AF = AE = BE = BC + EF = AD + EF -> EF = DF.  Since
DFE equals 80 degrees, FDE and DEF are both 50 degrees.  Since FDC is
30 degrees, CDE must be 20 degrees.

-- Craig Gentry

There is only one *answer*, but there are several (6?) ways of getting
it.  [...]  It is 20 degrees.  Show that AC * EC = DC * DC by computing
the lengths of these lines (OK, this uses both trigonometry and
algebra, which the original proposer asked me to avoid.  However, I
have seen a long solution that is pure geometry.)  So triangles ACD and
DCE are similar, and angle CDE = angle CAD = 20 degrees.

-- Barry Wolk <wolk@ccm.umanitoba.ca>

[ Editor: Craig Gentry reposted his solution in a slightly longer form,
and answered a question from Stan Hoffman <stanleyj@buffnet.net>: ]

>I mean why should CF = BC + EF ; the bases or the 2 triangles?

This is how we know CF = BC + EF:
  Since BCH is equilateral, CH = BC.  Since EFH is equilateral,
HF = EF.  Using these 2 equations we get CF = CH + HF = BC + EF.
  How did we know BCH and EFH were equilateral?  Well, we knew from the
get-go that angle EBC is 60 degrees.  It's easy to see that angle FCB
is also 60 degrees just by noticing symmetry (or if you wanted to make
a more formal proof, by proving that triangles FBC and EBC are
congruent to each other).  Since these 2 angles are both 60 degrees,
triangle BCH is equilateral.  We can prove that angles FEH and EFH are
both 60 degrees - making triangle EFH equilateral - by using the fact
that EF is parallel to BC.

-- Craig Gentry

[ Editor: Barry Wolk, on Gentry's solution: ]

That looks long, but it is actually quite a clever proof.  Here is a
small improvement.

Erase all parts of that diagram that extend outside triangle ABC.  So
the point G disappears, as does most of the first paragraph of that
solution.

Now H obviously lies on the bisector of angle A.  Then look at the two
triangles ABH and BAD.  They have the same angles, 10-20-150 in
degrees, so they are similar.  They also have the side AB in common, so
they are congruent.  Therefore, AD = BH.

Now proceed with the second paragraph of Craig Gentry's proof.

-- Barry Wolk <wolk@ccm.umanitoba.ca>

[ Editor: D lies on AB, so "triangle BAD" is degenerate; the triangle
with angles 20-10-150 is CAD (angle CAD = 20, ACD = 10), and it is
congruent to ABH because AC = AB, which gives AD = BH. ]

[ Editor: a wrong attempt, and its correction: ]

robson@LaBRI.U-Bordeaux.FR (Mike ROBSON) writes:
>DCE = 10 (subtraction)
>
>DBE = 20 (subtraction)
>
>B is the centre of the circumcircle of CED (angle subtended at centre
> = twice angle subtended at point of arc)

This is where your mistake occurred.  The center of the circumcircle of
CED is *not* B.  It is somewhere else on the circumcircle of BDE, or
more specifically on the circular arc from D to E through B, from all
of whose points (not just from B) the points D and E subtend an angle
of 20 degrees.

In other words, yes, if you have two points on a circle they subtend [Editor: word
missing in the original] twice the angle from the center as they do from another point on the
circle, but the implication does not always hold in the other
direction.

-- David A. Karr <karr@cs.cornell.edu>

[ Editor: a second version of the 20-degree triangle, posted in 1993 by
cta.larc@nasamail with a diagram that was not quoted.  From the solution
below, D is on AB and E on AC, and the angle sought is BED. ]

    Given: BAC = 20, ABC=ACB=80, BCD=50, and EBC=60.

I could swear I saw the 20-degree isoceles triangle here about three
years ago, but I couldn't find it in the archive either ...

    Assumptions:  x is the point where BE intersects DC.
                  y is the point where a line perpendicular
                  to BA intersects BE when drawn through D [...]

So angle ABE = 20, DYE = exterior angle of triangle BDY = 20 + 90 = 110

[skip the derivations of 12 or 13 other angles ...]

    WALA    :  BED=180-DYE-EDY=30.

When did we figure out angle EDY?

Alternative solution:

1. Angle BDC = 50 = BCD, so triangle BCD is isoceles with BD = BC.
2. Place F on the line between A and C so angle FBC = 20.
3. Then triangle BCF is isoceles with BF = BC.
4. Angle FBE = 40 = angle FEB so triangle BEF is isoceles with FE = BF.
5. From 1. and 3., BD = BF, angle DBF = 60, triangle BDF is equilateral.
6. From 4. and 5., FE = FD, so angle DEF = angle EDF.
7. Angle DFE = 180 - BFD - BFC = 180 - 60 - 80 = 40.
8. Angle DEF = (180 - 40)/2 = 70.
9. Angle BED = 70 - 40 = 30.

-- David Karr <karr@cs.cornell.edu>

In article <2g7854INNl72@NEBULA.SYSTEMSZ.CS.YALE.EDU>, Rajiv Mirani
<mirani-rajiv@cs.yale.edu> writes:

>Draw a triangle ABC with angle ABC = angle ACB = 80 degrees.  Draw a
>line BD meeting AC at D such that angle DBC = 50 degrees.  Similarly
>draw a line CE meeting AB at E such that angle ECB = 60 degrees.  Find
>angle DEC (preferably without using trigonometry).

This problem is a special case of a more general type.  Triangles with
rational angles in the given places have been called Adventitious
triangles.  I can probably dig up some references given time.  Here's
another way of getting the answer:

Take G in CA so that CG = CE.  Produce EG to H making GH = BC.  Join
AH.  Then angle AEH is easily seen to be 60 (all angles in degrees) and
AGH = 80.

AG = AC-CG = AC-CE = AC-AE = AB-AE = BE.  So triangles AGH and ECB are
congruent.  So angle AHG = angle ECB = 60.  So triangle AHE is
equilateral.  So EH = EG+GH = CG = CD+DG.  Hence EG=DG.  So angle GED =
50 and DEG = 30.

And here's another similar problem (from the 1916 [sic] Scholarship
Exam, Peterhouse and Sidney, Cambridge University):

On opposite sides of a base BC are described two triangles ABC, BCD
such that angle ABC = 30, angle ACB = 80, angle DBC=angle DCB = 50.
Show that if AD is drawn, angle DAC = 30 and find the other angles in
the figure.

Yet another is:

Take a quadrilateral ABCD with angles ABD=30, DBC=20, ADB=40 and
BDC=60.  Join AC and find all the angles.

-- Tony Davie <ad@dcs.st-and.ac.uk>

[ Editor: another solution of the second version above (BCD = 50,
CBE = 60, find BED = 30), sent in 1995 in reply to a query to the
rec.puzzles oracle that asked for an exact proof "(ie no trig)". ]

I have reoriented the figure and provided some labels.  I will not
attempt to draw diagonal lines using ASCII graphics; rather, I will
just indicate approximately where the points are and you will need to
draw the lines yourself.

We are given:
        isosceles triangle ABC with AB = AC                     (1)
        angle BAC = 20 degrees                                  (2)
        point D on AB with angle BCD = 50 degrees               (3)
        point E on AC with angle CBE = 60 degrees               (4)

We are to find:
        angle BED (in degrees).

                A

                *



                    * E


      D *






B *--------------------------* C

By (1), the base angles ABC and ACB are equal.  The third angle is
known by (2), and therefore the base angles satisfy

        angle ABC = angle ACB = 80 degrees.                     (5)

Knowing two of the angles in triangle BCD by (3) and (5), we may deduce
that the third is

        angle CDB = 50 degrees.                                 (6)

From (3) and (6), triangle BCD is an isosceles triangle with

        BC = BD.                                                (7)

With B as a center and BC as a radius, draw an arc meeting AC at O.
Then

        BC = BO                                                 (8)

and BCO is an isosceles triangle having base angles BCO = BOC.  The
isosceles triangle ABC shares a base angle with BCO, and therefore the
two triangles are similar.  This means the corresponding angles BAC and
CBO are equal, and by (2) this gives

        angle CBO = 20 degrees.                                 (9)

Consider triangle BOD.  From (7) and (8) we know that

        BD = BO,                                                (10)

and we also know from (5) and (9) that

        angle DBO = angle DBC - angle CBO
                  = angle ABC - angle CBO
                  = 80 degrees - 20 degrees
                  = 60 degrees.                                 (11)

From (10) and (11) it follows that BOD is an equilateral triangle, and
therefore

        OD = OB                                                 (12)

and

        angle BOD = 60 degrees.                                 (13)

By (12) we may draw a circle with center at O and passing through B and
D.  Extend BO to meet the circle at F, forming a diameter BOF.

                    * E


      D *





                                                      * F

                            * O

B *--------------------------* C

Consider triangle BCE.  Knowing two of its angles from (4) and (5), we
may deduce that the third angle is

        angle BEC = angle BEO = 40 degrees.                     (14)

From (4) and (9) we have

        angle OBE = angle CBE - angle CBO
                  = 60 degrees - 20 degrees
                  = 40 degrees.                                 (15)

By (14) and (15), triangle BOE is isosceles, with OB = OE, and
therefore E is on the circle BDF with OE a radius.

Draw FD.  The angle BFD subtends the central angle BOD, and therefore
the angle at F is half the central angle, which by (13) is

        angle BFD = 1/2 * angle BOD = 30 degrees.               (16)

Finally, angles BED and BFD are two angles with their vertices on the
circle BDEF which subtend the same central angle, implying that

        angle BED = angle BFD.                                  (17)

From (16) and (17), we have angle BED = 30 degrees, as was required.

-- Dave Seaman <ags@seaman.cc.purdue.edu>