in a right angled- triangle ABC with B =90, D and E are two points on BC such that BD=DE=EC prove that 8AE2=3AC2+5AD2
Answers
Answered by
0
Answer:
Given right DABC, right angled at B
D & E are points of trisection of the side BC.
Let BD=DE=EC=k
Hence we get BE=2k & BC=3k
In ΔABD by Pythagoras theorem we get
AD
2
=AB
2
+BD
2
AD
2
=AB
2
+k
2
Similarly, in ΔABE we get
AE
2
=AB
2
+BE
2
Hence AE
2
=AB
2
+(2k)
2
=AB
2
+4k
2
and
AC
2
=AB
2
+BC
2
=AB+(3k)
2
AC
2
=AB
2
+9k
2
Consider 3AC
2
+5AD
2
=3(AB
2
+9k)+5(AB
2
+4k
2
)
=8AB
2
+32k
2
=8(AB
2
+4k
2
)
∴3AC
2
+5AD
2
=8AE
2
.
Step-by-step explanation:
Attachments:
Similar questions