Question

in the diagram below, ac bisects ∠bad, be bisects ∠abd, m∠fed = 97° and m∠efc = 108°. find m∠fcd. you may assume lines that appear straight are straight but the figure is not otherwise drawn to scale.

Explanation:

Step1: Find vertical - angle

Since $\angle EFC$ and $\angle AFB$ are vertical angles, $\angle AFB=\angle EFC = 108^{\circ}$.

Step2: Calculate $\angle FAB+\angle FBA$

In $\triangle ABF$, using the angle - sum property of a triangle ($\angle FAB+\angle FBA+\angle AFB = 180^{\circ}$), we have $\angle FAB+\angle FBA=180^{\circ}-\angle AFB = 180 - 108=72^{\circ}$.

Step3: Use angle - bisector properties

Let $\angle BAC=\angle CAD = x$ and $\angle ABE=\angle EBD = y$. Then $2x + 2y=72^{\circ}$, so $x + y = 36^{\circ}$.

Step4: Consider $\triangle EFD$

We know $\angle FED = 97^{\circ}$. In $\triangle EFD$, let's consider the relationship between the angles in the overall figure.
We use the property that the exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles.
Let's consider the angles around point $F$.
We know that $\angle EFC = 108^{\circ}$, so its supplementary angle $\angle EFB=72^{\circ}$.
In $\triangle EFB$, we know part of the angle relationships.
Since $\angle FED = 97^{\circ}$, in $\triangle EFD$, we consider the angle at $F$ in the context of the bisected angles.
We know that the sum of angles in a triangle is $180^{\circ}$.
We use the fact that the angles formed by the bisectors and the given angles are related.
Let's consider the exterior - angle of $\triangle ABF$.
The exterior - angle of $\triangle ABF$ at $F$ is related to the angles in $\triangle EFD$.
We know that $\angle EFC = 108^{\circ}$.
We use the property that the sum of angles around a point is $360^{\circ}$.
Since $\angle EFC = 108^{\circ}$, and we know $\angle FED = 97^{\circ}$.
In $\triangle EFD$, we find the third - angle $\angle EDF$ in $\triangle EFD$ as $180-(97 + \angle EFD)$.
We know that $\angle EFD$ is related to the angles formed by the bisectors.
Since $\angle EFC = 108^{\circ}$, we consider the angle relationships in the figure.
We know that $\angle AFB=\angle EFC = 108^{\circ}$.
In $\triangle ABF$, we have $\angle FAB+\angle FBA = 72^{\circ}$.
Since $AC$ bisects $\angle BAD$ and $BE$ bisects $\angle ABD$.
We consider the fact that the sum of angles in $\triangle ABD$ and the relationships between the sub - triangles.
We know that $\angle EFC$ and $\angle EFD$ are supplementary. So $\angle EFD=72^{\circ}$.
In $\triangle EFD$, using the angle - sum property of a triangle ($\angle FED+\angle EDF+\angle EFD = 180^{\circ}$), we find $\angle EDF=180-(97 + 72)=11^{\circ}$.
Let's consider the angle $\angle FCD$.
We use the property of angle - bisectors and the exterior - angle of a triangle.
The exterior - angle of $\triangle ACD$ (related to the angles formed by the bisectors) and the given angles.
We know that $\angle EFC$ is an exterior - angle of $\triangle FCD$.
By the exterior - angle property of a triangle ($\angle EFC=\angle FCD+\angle FDC$).
We know $\angle EFC = 108^{\circ}$ and we found $\angle FDC = 11^{\circ}$.
So $\angle FCD=\angle EFC-\angle FDC$.

Step5: Calculate $\angle FCD$

$\angle FCD=108 - 11=97^{\circ}$

Answer:

$97^{\circ}$