Advertisements
Advertisements
प्रश्न
In ΔABC, D is a point in the interior of the triangle. Prove that DB + DC < AB + AC.
Advertisements
उत्तर
In the ΔABC,
AB + AC > BC .....(∵ Sum of the two sides of triangle is always greater than third side.) ....(i)
Also, in the ΔBDC,
BD + DC > BC ....(ii)
Dividing (i) by (ii),
`"AB + AC"/"BD + DC" > "BC"/"BC"`
`"AB + AC"/"BD + DC" > 1`
AB + AC > BD + DC
i.e. BD + DC < AB + AC.
APPEARS IN
संबंधित प्रश्न
In the given figure sides AB and AC of ΔABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.
ABC is a triangle. Locate a point in the interior of ΔABC which is equidistant from all the vertices of ΔABC.
How had the position of women improved in our country since independence ? Explain with examples.
From the following figure, prove that: AB > CD.
In the following figure, write BC, AC, and CD in ascending order of their lengths.
"Issues of caste discrimination began to be written about in many printed tracts and essays in India in the late nineteenth century." Support the statement with two suitable examples.
Name the smallest angle in each of these triangles:
In ΔABC, AB = 6.2cm, BC = 5.6cm and AC = 4.2cm
In the given figure, ∠QPR = 50° and ∠PQR = 60°. Show that : PN < RN
In ΔABC, BC produced to D, such that, AC = CD; ∠BAD = 125° and ∠ACD = 105°. Show that BC > CD.
In ΔPQR is a triangle and S is any point in its interior. Prove that SQ + SR < PQ + PR.
