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
3.9k views
in Triangles by (65.2k points)
closed by

Let ABC be a triangle and P be an interior point, prove that AB + BC + CA < 2(PA + PB + PC).

1 Answer

+1 vote
by (65.5k points)
selected by
 
Best answer

In a triangle, sum of any two side is grater than the third side 

\(\therefore\) In ΔPBA, AB < PA + PB ….(i) 

In ΔPBC, BC < PB + PC ….(ii) 

In ΔPCA, AC < PC + PA ….(iii) 

By adding (i), (ii) and (iii) 

AB + BC + AC < PA + PB + PB + PC + PC + PA 

AB + BC + AC < 2 (PA+ PB + PC).

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

...