Question

It is given that \(\triangle ABC \sim \triangle EDF\). Which of the following is not true?

• A. \(\frac{\text{Perimeter of } \triangle ABC}{\text{Perimeter of } \triangle EDF} = \frac{AB}{ED}\)
• B. \(\frac{AB}{ED} = \frac{AC}{EF}\)
• C. \(\angle A = \angle D, \angle C = \angle F\)
• D. \(\frac{AB + BC}{AC} = \frac{ED + DF}{EF}\)

Hint:

In similarity notation \(\triangle XYZ \sim \triangle PQR\), position 1 matches position 1 (\(X \leftrightarrow P\)), position 2 matches position 2 (\(Y \leftrightarrow Q\)), and so on. Use this to write ratios correctly every time.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In similar triangles, the order of vertices determines the corresponding parts. For \(\triangle ABC \sim \triangle EDF\):
Corresponding angles are: \(\angle A = \angle E\), \(\angle B = \angle D\), \(\angle C = \angle F\).
Corresponding side ratios are: \(\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} = k\).
Step 2: Key Formula or Approach:
Check each option against these rules.
Step 3: Detailed Explanation:
(A) True: The ratio of perimeters is equal to the ratio of corresponding sides.
(B) True: \(\frac{AB}{ED} = \frac{AC}{EF}\) is a correct pair of corresponding sides.
(C) False: According to the order of vertices, \(\angle A\) corresponds to \(\angle E\). Therefore, \(\angle A = \angle E\) is true, but \(\angle A = \angle D\) is generally not true.
(D) True: Since \(\frac{AB}{ED} = \frac{BC}{DF} = \frac{AC}{EF} = k\), then \(AB = k \cdot ED\), \(BC = k \cdot DF\), and \(AC = k \cdot EF\).
Substituting these: \(\frac{k \cdot ED + k \cdot DF}{k \cdot EF} = \frac{k(ED + DF)}{k \cdot EF} = \frac{ED + DF}{EF}\). This holds true.
Step 4: Final Answer:
The statement (C) is not true.