Find angle EDC in congruent triangles problem | 70° solution

Image Description

The image displays two right-angled triangles, $\triangle ABF$ and $\triangle EDG$, positioned on a horizontal line segment $AE$. There are right angles at vertices $A$ and $E$. The hypotenuses $BF$ and $DG$ intersect at point $C$, creating a smaller triangle $CGF$ in the center. The problem states that $\triangle ABF \cong \triangle EDG$ and provides the measure of $\angle BCG = 40^{\circ}$.

Answer

The measure of $\angle EDC$ is $70^{\circ}$. This is determined by using the properties of congruent triangles to establish an isosceles triangle $\triangle CGF$, which allows us to calculate the interior angles of $\triangle EDG$.

Explanation

1. Identify corresponding parts of congruent triangles The problem states $\triangle ABF \cong \triangle EDG$. According to CPCTC (Corresponding Parts of Congruent Triangles are Congruent), the corresponding angles must be equal. Specifically, the angle at $F$ in the first triangle corresponds to the angle at $G$ in the second triangle. $\angle AFB = \angle EGD$ This formula shows that the base angles of the two overlapping triangles are identical.

2. Determine the nature of triangle CGF Since $\angle CFG$ (which is the same as $\angle AFB$) is equal to $\angle CGF$ (which is the same as $\angle EGD$), $\triangle CGF$ is an isosceles triangle with $CG = CF$. We also know that $\angle BCG$ and $\angle CGF$ are related through the Exterior Angle Theorem or vertical angles. However, let's use the given vertical angle: $\angle FCG = \angle BCD$ (if lines were straight) is not applicable here. Instead, observe $\triangle BCG$. By the Exterior Angle Theorem for $\triangle CGF$: $\angle BCF = \angle CGF + \angle CFG$ Actually, a simpler path: $\angle BCG$ is given as $40^{\circ}$. In $\triangle CGF$, the exterior angle at vertex $C$ (which is $\angle BCG$) is equal to the sum of the two opposite interior angles. $\angle BCG = \angle CFG + \angle CGF$ The exterior angle of a triangle is equal to the sum of the two remote interior angles.

3. Calculate the base angles Since $\angle CFG = \angle CGF$, let both be represented by $x$. $40^{\circ} = x + x$ $40^{\circ} = 2x$ $x = 20^{\circ}$ This means $\angle EGD = 20^{\circ}$. ⚠️ This step is required on exams to prove the relationship between the exterior angle and the base angles.

4. Solve for the target angle in triangle EDG Now consider the right triangle $\triangle EDG$. We know $\angle DEG = 90^{\circ}$ and we just found $\angle EGD = 20^{\circ}$. The sum of angles in a triangle is $180^{\circ}$. $\angle EDG = 180^{\circ} - 90^{\circ} - 20^{\circ}$ The interior angles of any triangle must sum to 180 degrees. $\angle EDG = 70^{\circ}$ This calculation gives us the total angle at vertex $D$. Since $C$ lies on the line segment $DG$, $\angle EDC$ is the same as $\angle EDG$.

Final Answer

$\boxed{70^{\circ}}$

Common Mistakes

• Assuming symmetry incorrectly: Students often assume $\angle BCG$ is equal to $\angle CGF$, leading to an incorrect base angle of $40^{\circ}$ and a final answer of $50^{\circ}$.
• Misidentifying corresponding angles: Ensure you match the vertices based on the congruence statement ($\triangle ABF \cong \triangle EDG$). If you match $B$ with $G$ instead of $D$, the angle relationships will fail.