Question

In $\triangle DEF$ shown, points A, B, and C are taken on DE, DF, and EF respectively such that $EC = AC$ and $CF = BC$. If $\angle D = 40^\circ$, then $\angle ACB = $?

• A. 140
• B. 70
• C. 100
• D. None of these

Hint:

Mark equal sides and use isosceles triangle properties to deduce equal angles for angle chasing.

Answer 1

The Correct Option is B

$EC = AC$ implies $\triangle AEC$ is isosceles with $\angle EAC = \angle ACE$. Similarly, $CF = BC$ implies $\triangle BCF$ is isosceles with $\angle FBC = \angle BCF$. Angle chasing in the geometry shows $\angle ACB = 70^\circ$.