Question

In triangle $ABC$, $AB = AC$ and $\angle A = 100^\circ$. $D$ is the point on $BC$ such that $AC = DC$, and $F$ is the point on $AB$ such that $DF$ is parallel to $AC$. Determine $\angle DCF$.

• A. 30^\circ
• B. 40^\circ
• C. 60^\circ
• D. 70^\circ

Answer 1

60^\circ

Answer 2

1. In $\triangle ABC$, $AB = AC$ and $\angle A = 100^\circ$. Thus, $\angle B = \angle C = (180^\circ - 100^\circ)/2 = 40^\circ$.
2. In $\triangle ADC$, $AC = DC$ and $\angle ACD = \angle ACB = 40^\circ$. Thus, $\angle DAC = \angle ADC = (180^\circ - 40^\circ)/2 = 70^\circ$.
3. Since $DF \parallel AC$, $\angle FDB = \angle ACB = 40^\circ$ (corresponding angles).
4. Also, since $DF \parallel AC$, $\angle FDC = \angle ACD = 40^\circ$ (alternate interior angles).
5. In $\triangle DFC$, $\angle DFC = 180^\circ - \angle FDB - \angle B = 180^\circ - 40^\circ - 40^\circ = 100^\circ$. (This step is incorrect. $\angle DFC$ is not an angle in $\triangle DFC$ directly from $\angle FDB$ and $\angle B$. Instead, consider $\angle AFD$. Since $DF \parallel AC$, $\angle BFD = \angle BAC = 100^\circ$ (corresponding angles). Then $\angle AFD = 180^\circ - 100^\circ = 80^\circ$.)
6. In $\triangle DFC$, $\angle DCF = 180^\circ - \angle FDC - \angle DFC = 180^\circ - 40^\circ - 80^\circ = 60^\circ$.