Question

△ ABC is an acute angled triangle. DE is drawn parallel to BC as shown. Which of the following are always true?

i) △ABC∼△ ADE

ii) ADBD=AEEC

iii) DE=BC2

(i), (ii) and (iii)

(ii) and (iii) only

(i) and (ii) only

Only (i)

Solution

The correct option is **C**

(i) and (ii) only

Consider △ADE and △ABC

Since DE || BC,

ADBD=AEEC

(by basic proportionality theorem)

∠BAC=∠DAE (common in both the triangles)

∴△ADE∼△ABC (by SAS similarity criterion)

⇒ADAB=DEBC=AEAC

(corresponding sides of similar triangles are in same ratio)

Now, DE=BC2 only if

⇒ADAB=DEBC=AEAC=12

i.e. D and E should be the midpoints of AB and AC respectively.

So this may not be true always.

Suggest corrections

0 Upvotes

Similar questions

View More...

People also searched for

View More...

- About Us
- Contact Us
- Investors
- Careers
- BYJU'S in Media
- Students Stories - The Learning Tree
- Faces of BYJU'S – Life at BYJU'S
- Social Initiative - Education for All
- BYJU'S APP
- FAQ
- Support

© 2021, BYJU'S. All rights reserved.