Question

Rectangle ABCD is shown with diagonal AC drawn. Point F is on AC and point E on DC such that FE is perpendicular to DC. Prove triangle ABC is similar to triangle CEF

Answer 1

Triangle ABC is similar to triangle CEF.

Explanation:

Diagram is inserted for the reference.

ABCD is a rectangle.

ABC is a right angled triangle because all the angles of the rectangle are 90◦ - (a)

CEF is a right angled triangle because FE is perpendicular to DC – (b)

In triangles ABC and CEF,

1. Angle ABC = Angle CEF = 90◦ (Both are right angles from a and b)

2. Angle BCA = Angle EFC (Alternate angles on parallel lines are equal on intersection)

Hence using Similarity property of AA (Angle, Angle), Triangle ABC and CEF are similar.

Answer 2

ΔABC and ΔCEF are similar triangles by the AA similarity theorem.

What is the AA Similarity Theorem?

The angle-angle similarity theorem (AA) states that when two triangles have two pairs of corresponding congruent angles, both triangles are similar triangles.

In ΔABC and ΔCEF, we have the following:

Two pairs of corresponding congruent angles - ∠FEC ≅ ∠ABC (right angles) and ∠FCE ≅ ∠BAC

Therefore, ΔABC and ΔCEF are similar triangles by the AA similarity theorem.

Learn more about the AA similarity theorem on:

https://brainly.com/question/2166570