Question

in the figure, $overrightarrow{ca}$ and $overrightarrow{ce}$ are opposite rays, $overrightarrow{ch}$ bisects $angle gcd$, and $overrightarrow{gc}$ bisects $angle bgd$. if $mangle gcf$ is $12^{circ}$ less than $mangle bcg$, what is $mangle fcd$? $mangle fcd = square^{circ}$

Explanation:

Step1: Define angle - bisector properties

Since $\overrightarrow{CH}$ bisects $\angle GCD$, then $\angle GCF=\angle FCD$. Since $\overrightarrow{GC}$ bisects $\angle BGD$, let $m\angle BCG = x$. Then $m\angle GCF=x - 12^{\circ}$.

Step2: Use angle - relationships

We know that $\angle BCG+\angle GCF+\angle FCD = 180^{\circ}$ (because $\overrightarrow{CA}$ and $\overrightarrow{CE}$ are opposite rays, so $\angle ACE = 180^{\circ}$). Substitute $\angle GCF=x - 12^{\circ}$ and $\angle FCD=x - 12^{\circ}$ (since $\angle GCF=\angle FCD$) and $\angle BCG = x$ into the equation: $x+(x - 12^{\circ})+(x - 12^{\circ})=180^{\circ}$.

Step3: Solve the equation for $x$

Combine like - terms: $3x-24^{\circ}=180^{\circ}$. Add $24^{\circ}$ to both sides of the equation: $3x=180^{\circ}+24^{\circ}=204^{\circ}$. Divide both sides by 3: $x = 68^{\circ}$.

Step4: Find $m\angle FCD$

Since $m\angle FCD=x - 12^{\circ}$, substitute $x = 68^{\circ}$ into the equation. Then $m\angle FCD=68^{\circ}-12^{\circ}=56^{\circ}$.

Answer:

$56$