i.e,. BD=DC
To prove : AB^2+AC^2=2AD^2+2BD^2
Construction: Draw AE _l_ BD
Proof: in ∆ABE, L AEB=90°By Pythagoras theorem
AB^2=AE^2+BE^2 ----------->1
In ∆ACE, L AEC =90°
AC^2=AE^2+EC^2. ------------>2
In ∆AED, L AED =90°
By Pythagoras theorem,
AD^2=AE^2+ED^2
2AD^2=2AE^2+2ED^2. ---------->3
Adding 1&2 we get
AB^2+AC^2=2AE^2+BE^2. --------->4
Subtract 4&3 we get
AB^2+AC^2-2AD^2=2AE^2+BE^2+EC^2 -2AE^2-2ED^2
AB^2+AC^2-2AD^2=BE^2+EC^2-2ED^2
=(BD-ED)^2+(ED+DC)^2- 2ED^2
=BD^2-2BD•ED+ED^2
+ED^2+2ED•DC+DC^2-2ED^2
=BD^2-2BD•ED+2ED•DC
+DC^2
=BD^2-2BD•ED+2ED•BD
+BD^2
[=2BD^2]
AB^2+AC^2-2AD^2=2BD^2
[ AB^2+AC^2=2BD^2+2AD^2]
Very thinking ЁЯдФ
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