Question

Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third angle.

Answer 1

Answer:

Given : Triangle ABC in which AD is the median.

To prove:AB+AC>2AD

Construction :

Extend AD to E such that AD=DE .

 Now join EC.

Proof:

In ΔADB and ΔEDC

AD=DE[ By construction]

D is the midpoint BC.[DB=DB]

ΔADB=ΔEDC [vertically opposite angles]

Therefore Δ ADB ≅  ΔEDC [ By SAS congruence criterion.]

--> AB=ED[Corresponding parts of congruent triangles ]

In ΔAEC,

AC+ED> AE [sum of any two sides of a triangle is greater than the third side]

AC+AB>2AD[AE=AD+DE=AD+AD=2AD and ED=AB]

Hence proved 

Answer 2

Given : Triangle ABC in which AD is the median.

To prove:AB+AC>2AD

Construction :

Extend AD to E such that AD=DE .

 Now join EC.

Proof:

In ΔADB and ΔEDC

AD=DE[ By construction]

D is the midpoint BC.[DB=DB]

ΔADB=ΔEDC [vertically opposite angles]

Therefore Δ ADB ≅  ΔEDC [ By SAS congruence criterion.]

--> AB=ED[Corresponding parts of congruent triangles ]

In ΔAEC,

AC+ED> AE [sum of any two sides of a triangle is greater than the third side]

AC+AB>2AD[AE=AD+DE=AD+AD=2AD and ED=AB]

Hence proved