ABC is a triangle and D is any point in its interior
show that BD+DC
Answers
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.
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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.