Question

in the adjoining figure ABC is a triangle and D is any point in its interior. show that (BD+DC) < (AB+AC).​

Answer 1

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in the adjoining figure ABC is a triangle and D is any point in its interior. show that (BD+DC) < (AB+AC).

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Produce BD to meed AC at E.

We know that in a triangle, the sum of any two sides is always greater than the third side,

In ∆ABE,we have

AB + AE > BE ⇒ AB + AE > BD + DE.⠀⠀...(i)

In ∆CDE,we have

DE + EC > DC.⠀⠀⠀⠀..(ii)

From (i) and (ii), we get

AB + AE + DE + EC > BD + DE + DC

⇒AB + (AE + EC) > BD + DC

⇒(AB + AC) > (BD + DC).

Hence,(BD + DC) < (AB + AC).

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Answer 2

Question - the adjoining figure ABC is a triangle and D is any point in its interior. show that (BD+DC) < (AB+AC).

Solution - Produce BD to meed AC at E.

We know that in a triangle, the sum of any two sides is always greater than the third side,

In ∆ABE,we have

AB + AE > BE ⇒ AB + AE > BD + DE.⠀⠀...(i)

In ∆CDE,we have

DE + EC > DC.⠀⠀⠀⠀..(ii)

From (i) and (ii), we get

AB + AE + DE + EC > BD + DE + DC

⇒AB + (AE + EC) > BD + DC

⇒(AB + AC) > (BD + DC).

Hence,(BD + DC) < (AB + AC)...

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