AB and AC are equal legs in ABC. Point D lies on AB such that CD=CB. If ADC=114, what is ACD in degrees?
https://latex.artofproblemsolving.com/0/0/2/00269e96583c7f0dbf6e4cd0735f4b37ed1511f2.png
If ADC = 114, then BDC = 180-114 = 66.
Since BC is congruent to DC, angle BDC is congruent to angle DBC, which is also 66 degrees
The angle measures of triangle BCD must add up to 180 degrees, so 66+66 = 132 and 180 - 132 = 48. So angle BCD = 48.
Now, side AB is congruent to side AC, so angle ABC is equal to angle ACB.
Can you solve it from there? (hint - angle ABC is also angle DBC.)
You are very welcome!
:P
Since CD = CB.....then angles DBC and BDC are equal
And since angle ADC = 116°.....then angle BDC = 180 - 116 = 64°
So angle DBC = 64°
Then angle DCB = 180 - 2(64) = 180 - 128 = 52°
And since AB = AC.....then angles ABC and ACB are equal
And angle ABC = angle DBC
So angle ACB = 64°
And angle ACB = angle DCB + angle ACD
So
64 = 52 + angle ACD subtract 52 from both sides
12° = angle ACD
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