Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
221 views
in Triangles by (51.8k points)
closed by

ΔABC is a right angled triangle. ∠B = 90° a perpendicular line from B to AC intersect AC at D. Prove that ΔABC, ΔADB and ΔBCD are similar to each other.

1 Answer

+1 vote
by (49.3k points)
selected by
 
Best answer

ΔABC = ΔABD 

[∠ABC = ∠ADB = 90°, ∠ACB = ∠ABD = 90 – X] 

∠A (common) 

similarly ΔABD ~ ΔBDC 

In ΔABC, consider 

∠A = X° 

∠B = 90° 

∠A + C = 90°, ∠C = 90 – X 

In ΔABD, ∠ADB = 90° 

Therefore ∠A = X°, ∠ABC = 90° 

∠A + ∠ABD = 90° 

∠ABD = 90 – ∠A = 90 – X; 

In ΔBCD, ∠BDC = 90° 

∠C = (90 – X); ∠C + ∠DBC = 90°; 

∠DBC = 90 – (90 – X) = X° 

Hence angles of three triangles are 90°, X° and 90° – X°.

Hence three triangles are similar to each other.

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...