In isosceles $\triangle abc$ (with $ab = ac$), point $d$ lies on $ab$ such that $cd = cb$. if $\angle adc = 115^\circ$, what is $\angle acd$ (in degrees)?

1. Angles ADC and CDB are supplementary, thus

m∠ADC+m∠CDB=180°.

Since m∠ADC=115°, you have that m∠CDB=180°-115°=65°.

2. Triangle BCD is isosceles triangle, because it has two congruent sides CB and CD. The base of this triangle is segment BD. Angles that are adjacent to the base of isosceles triangle are congruent, then

m∠CDB=m∠CBD=65°.

The sum of the measures of interior angles of triangle is 180°, therefore,

m∠CDB+m∠CBD+m∠BCD=180° and

m∠BCD=180°-65°-65°=50°.

3. Triangle ABC is isosceles, with base BC. Then

m∠ABC=m∠ACB.

From the previous you have that m∠ABC=65° (angle ABC is exactly angle CBD). So

m∠ACB=65°.

4. Angles BCD and DCA together form angle ACB. This gives you

m∠ACB=m∠ACD+m∠BCD,

m∠ACD=65°-50°=15°.

15°.

i dont know

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