## Another Isosceles Triangle

Consider a triangle ABC with side AB=AC, and angle BAC = 20 degrees. D is a point on side AC. AD=BC. Find angle DBC.

Source: Internet newsgroup rec.puzzles, February 8, 1996.

Solutions were received from Philippe Fondanaiche, Ajay Singh, Jimm Chng Gim Hong, Radu Ionescu, Dane Brooke, Jozef Hanenberg, Graeme McRae, Kenneth Suprin, Mete Kart, Claudio Baiocchi, Hareendra Yalamanchili, John Hewson, Jayavel Sounderpandian, and Denis Borris. The angle measures 70 degrees. There were many different proofs and I've included a few here.
Let be E on BC and BE=EC.
Let be F on AE so that triangle FBC is equilateral.
angleFAD=angleDBF= 10 degrees ==> angleDBC=10 +60= 70 degrees
From Dane Brooke:
```We have triangle ABC with angle BAC 20 degrees.
Construct triangle ABE with angle ABE 20 degrees and BE = AB and BE
intersecting AC at F.
(These triangles are identical as are AFE and BFC.)
Bisect angle ABE and call the intersection of the bisector with AC the point
D.
Observe that triangles BDE and BDA are also identical.
Now we have angles EAD and AED equal to each other and to 60 degrees, whence
angle ADE must also be 60 degrees.
Evidently BC = AE = AD. How lucky! My point D coincides with your point D!
Since angle ABD is 10 degrees and angle ABC is 80 degrees, angle DBC is
evidently 70 degrees.
```

From Graeme McRae:
```For simplicity, let sides AB=AC have length 1.  Then side BC has length
2 sin(10).    (If point E is the midpoint of BC then triangle AEC is a
right with angle EAC equal to 10 degrees.)

Let x = angle ABD.  Then angle BDA is 180-x-20.  By the law of sines,

sin(180-x-20)/1 = sin(x+20)/1 = sin(x) / (2 sin 10)

x=10 is a solution to this because

sin(10+20)/1 = 1/2 = sin(10)/ (2 sin 10)

Since angle ABC is 80 and angle ABD is 10, angle DBC is 70.
```

From Kenneth Suprin:
```Use the Law of Sines on Triangle ABC.  This tells me BC =.347296*AB.
Angle BAC = 20 degrees
Angle BCA = 80 degrees = Angle ABC

I then set AB = 1 for ease in computation.

I then looked at Triangle ABD.

AB= 1
BC = .347296...

Angle ABD = x
Angle BDA = 180 -20 -x = 160 - x

Once again I used the law of sines.

sin (x)/sin (160-x) = .347296/1

Now solve for x.

I also used the trig identity: sin(x-y) = sin(x)*cos(y) - cos(x)*sin(y)
where y = 160 in this case.

After some algebra, you get tan (x) = .17632...

Take the arctan to find out that Angle ABD =10 degrees.

Now Angle ABD + Angle DBC = angle ABC

From above: angle ABD =10 degrees, angle ABC = 80 degrees, so angle DBC must
be 70 degrees.
```

From John Hewson:
```Let AB = AC = 1 and BC = 2 sin 10.
To compute angle DBC first find BD knowing BC = 2 sin 10, DC = 1 - 2 sin 10
and angle BCD = 80 degrees.  Then find angle DBC, in the same triangle,
using the constancy of the ratios of sides to the sines of opposite angles.

It remains to explain why angle DBC = (90 - angle BAC) degrees.
Let angle BAC = 2a degrees.      What value of a will make angle DBC = 90 -
2a degrees?
Angle BCD = 90 - a, so with DBC = 90 - 2a, angle BDC = 3a.
Let BC = 2x.   Then x = sin a, and DC = 1 - 2x.
The sides of the triangle DBC are proportional to the sines of the opposite
angles.   That is:
(sin 3a)/2x=(cos 2a)/(1 - 2x)
ie (sin 3a)/2sin a = (cos 2a)/(1 - 2 sin a)
ie (3 - 4 (sin a)^2)(1 - 2sin a) = 2(1 - 2(sin a)^2)
or 8(sin a)^3 - 6(sin a) + 1= 0
ie 2 sin 3a =1, 3a = 30 degrees, a = 10 degrees (the angle given in the
puzzle), angle BAC = 20 degrees, and DBC = 90 - 20 = 70 degrees.
```

For completeness, here is the solution posted on rec.puzzles with the original puzzle. I don't know who the author was.
```Find the point E such that EA = ED = AB, with E and B both on the same side
of line ADC. I will omit the easy proof of each of the following facts:

triangles EAD and ABC are congruent
angle AED = 20