Question

In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.

Answer 1

In ΔPQR,

PQ + PR > QR   ....(∵ Sum of two sides of a triangle is always greater than the third side.)     ....(i)

Also, in ΔSQR,

SQ + SR > QR ....(∵ Sum of two sides of a triangle is always greater than the third side.)     ....(ii)
Dividing (i) by (ii),

`"PQ + PR"/"SQ + SR" > "QR"/"QR"`

`"PQ + PR"/"SQ + SR" > 1`
PQ + PR > SQ + SR
i.e. SQ + SR < PQ + PR.