Question

A triangle ABC with sides BC =7cm, $\angle B={45}^{\circ },\angle A={105}^{\circ }$ is given. The image of constructing a similar triangle of $\Delta ABC$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $\Delta ABC$ is given below.

The condition used to draw A'C' parallel to AC:

• A. Corresponding angles are equal
• B. Alternate interior angles are equal
• C. Co-interior angles are supplementary
• D. Perpendicular bisector theorem

Answer 1

The correct option is A Corresponding angles are equal


Here we use the principle that when corresponding angles are equal, the lines are parallel.
$\angle A^{\prime }{C}^{\prime }B=\angle ACB\to$ AC||A'C'.