# In the given figure DB ⊥ BC, DE ⊥ AB and AC ⊥ BC, prove that ΔBDE ~ΔABC.

We have, DB BC and AC BC

B + C = 90° + 90°

B + C = 180°

BD || AC

⇒∠EBD = CAB (alternate angles)

Let us take BDE andABC

BED = ACB (each 90°)

EBD = CAB (alternate angles)

BDE ~ABC (by AA similarity criterion)

Hence Proved

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Basic Proportionality Theorem42 mins
A Peep into Pythagoras Theorem43 mins
R.D Sharma | Solve Exercise -4.2 and 4.3FREE Class
NCERT | Strong Your Basics of Triangles39 mins
RD Sharma | Imp. Qs From Triangles41 mins
Champ Quiz | Thales Theorem49 mins
NCERT | Concept Based Questions on Triangles31 mins
Quiz | Criterion of Similarity of Triangle45 mins
How to Ace Maths in NTSE 2020?36 mins
Know About Important Proofs in Triangles33 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses