Prove that the sum of any two sides of a triangle is greater than twice the median drawn to the third angle.
Answers
Answered by
0
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
Answered by
0
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
Similar questions