Thread: A triangle question

1. A triangle question

In a triangle ABC, ∠ABC=112 and AB=BC. D is a point on the side AB and E is a point on the side AC such that AE=ED=DB. Find the size of ∠BCD.

I have tried using a circle to solve this problem, but it doesn't seem to help. Help me explain it, please!

2. AB = BC, so triangle ABC then is isosceles.
Angle BAC = angle BCA = (1/2)(180 -112) = 34 degrees.

AE = ED, so triangle AED is isosceles too.
Angle ADE = angle BAC = 34 deg
Angle AED = 180 -34 -34 = 112 deg

Angle AEC is 180 deg.
Angle DEC = 180 -112 = 68 deg

This is the key to the problem: ED = DB. It must mean something.

Draw a line segment DF from point E to base AC such that DF is equal to ED.
Triangle EDF is isosceles.
Angle DFE = angle DEF = angle DEC = 68 deg
Angle EDF = 180 -68 -68 = 44 deg

Draw line segment BF.
DF = DB, so triangle BDF is isosceles.
Angle BDA is 180 deg. Angle DEF = 180 -44 - 34 = 102 deg
Angle DBF = angle DFB = (1/2)(180 -102) = 39 deg

Quadrilateral DBCA looks like a kite.
Any quadrilateral has 360 degrees for the sum of its interior angles.
(Interior angles = (n-2)(180) = (4-2)(180) = 360 deg.)
So, angle DFC = 360 -102 -112 -34 = 112 deg
That means angle DFC = angle DBC.
That means DBCA is a kite. That means DC is a diagonal, the longer diagonal and it bisects angle BCF.
Angle BCF = angle BCA = 34 deg
Therefore,
angle BCD = angle FCD = (1/2)(34) = 17 degrees. ----------answer.

------------------
Another way, after finding angle DFC = angle DBC = 112 deg.
By triangle congruence SsA, meaning the given angle is opposite the longer side.
SSA is not 100% true for congruence. But there are special cases of SSA where congruence can be proven. One of them is SsA. Its commonly called "SsA Congruence Theorem"
(You can search the internet for that.)

CD = CD
DB = DF
angle DBC = angle DFC
SsA, so triangle CDB = triangle CDF = (1/2)(quadrilateral DBCA)
That means angle BCF is bisected by line segment CD.
Therefore, angle BCD = (1/2)(34) = 17 degrees ---------answer.

3. Thank you tibcol, a million times! I left out some of the properties before, that's why I can't do it.

4. Me, I included more than enough. :-)
I was in a hurry to go to work this morning so I was not able to edit very well my answer above.

"....
Angle EDF = 180 -68 -68 = 44 deg

Draw line segment BF.
DF = DB, so triangle BDF is isosceles.
Angle BDA is 180 deg. Angle DEF = 180 -44 - 34 = 102 deg
Angle DBF = angle DFB = (1/2)(180 -102) = 39 deg

Quadrilateral DBCA looks like a kite.
Any quadrilateral has 360 degrees for the sum of its interior angles.
(Interior angles = (n-2)(180) = (4-2)(180) = 360 deg.)
So, angle DFC = 360 -102 -112 -34 = 112 deg
That means angle DFC = angle DBC.
That means DBCA is a kite. That means DC is a diagonal, the longer diagonal and it bisects angle BCF.
Angle BCF = angle BCA = 34 deg
Therefore,
angle BCD = angle FCD = (1/2)(34) = 17 degrees. ----------answer."

should not have been included anymore.
Or, I could have continued differently, like so:

"Draw a line segment DF from point E to base AC such that DF is equal to ED.
Triangle EDF is isosceles.
Angle DFE = angle DEF = angle DEC = 68 deg ...."
Angle DFC = 180 -68 = 112 deg -----same as angle DBC

Draw line segment CD.
Compare triangles CDB and CDF.
"CD = CD
DB = DF
angle DBC = angle DFC
SsA, so triangle CDB = triangle CDF = (1/2)(quadrilateral DBCA)
That means angle BCF is bisected by line segment CD.
Therefore, angle BCD = (1/2)(34) = 17 degrees ---------answer."

"...triangle congruence SsA, meaning the given angle is opposite the longer side, is a special case of SSA ---where congruence can be proven. It is commonly called "SsA Congruence Theorem"

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