ABC is a triangle and D is any point in its interior show that BD+DC<AB+AC please help me fast
Answers
Answer:
BD + DC < AB + AC
Step-by-step explanation:
Sum of two sides of triangle > Third side
in ΔABE
=> AB + AE > BE
=> AB + AE > BD + DE
in ΔDCE
DE + CE > DC
Adding both
AB + AE + DE + CE > BD + DE + DC
Cancelling DE from both sides
=> AB + AE + CE > BD + DC
AE + CE = AC
=> AB + AC > BD + DC
=> BD + DC < AB + AC
QED
Proved
Answer:
Proved that, BD + DC < AB +AC
Given
ABC is a triangle and D is any point in its interior.
To prove that, BD + DC < AB +AC
In ΔABE, AB + AE > BE
AB + AE > BE
Where, sum of any two sides of a triangle must be greater than the third side.
From the figure, BE = BD + DE
AB + AE > BE
AB + AE > BD + DE -----> ( 1 )
In ΔCDE, CE + ED > CD -----> ( 2 )
Now, add equation ( 1 ) and ( 2 )
AB + AE + CE + ED > BD + DE + CD
AB + AC + ED > BD + DE + CD ( AE + CE = AC )
AB + AC > BD + CD.
Hence, proved BD + DC < AB + AC.