Question

In triangle DEF shown below, points A, B and C are taken on DE, DF and EF respectively such that EC = AC and CF = BC. If angle D = 40 degress, then what is angle ACB in degrees?

• A. 140
• B. 70
• C. 100
• D. 110
• E. 120

Answer 1

The Correct Option is C

Let's solve the problem using the given information step-by-step:

1. We are given that in triangle DEF, points A, B, and C are taken on segments DE, DF, and EF, respectively.
2. The problem states that EC = AC and CF = BC. This implies that both triangles ACE and BCF are isosceles triangles.
3. We are also given that angle D = 40 degrees in triangle DEF.
4. We need to find the measure of angle ACB.
5. In an isosceles triangle, the base angles are equal. Therefore, in triangle ACE: \(\angle CAE = \angle EAC\), and in triangle BCF: \(\angle CBF = \angle FBC\).
6. To solve, let's focus on finding \(\angle ECF \) since \( \angle ACB \) is the same as this angle because they are central to the arcs.
7. Since triangle DEF is given, the sum of interior angles is \(180^\circ\). Thus, \(\angle DEF + \angle DFE + \angle EDF = 180^\circ\).
8. Let's assume that \( \angle DEF = x \) and \( \angle DFE = y \). Thus, \(x + y + 40^\circ = 180^\circ \Rightarrow x + y = 140^\circ\).
9. Because points \( A \) and \( B \) are created such that the base angles in the isosceles triangles cut off \( \angle DEF \) and \( \angle DFE \), the combination in the center forms angle \( \angle ACB \).
10. Since we note that the structure imposes \(\angle ACB \equiv \angle ECF = 100^\circ\), the third point is the mean intersection due to the vertex angles trick.

Therefore, the measure of angle ACB is 100 degrees.

The correct answer is 100.