The given figure shows a triangle ABC, in
which AB > AC. E is the mid-point of BC
and AD is perpendicular to BC.
Prove that : AB2 - AC2 = 2BC X ED
Answers
Answer:
In ΔABD,
AB
2
=BD
2
+AD
2
--(1)
In ΔADC,
AC
2
=CD
2
+AD
2
--(2)
Adding (1) and (2),
AB
2
+AC
2
=2AD
2
+BD
2
+CD
2
...........(8)
BD
2
=BE
2
+ED
2
+2BE.ED --(3)
CD=EC−ED
CD
2
=EC
2
+ED
2
−2EC.ED --(4)
Adding (3) and (4),
BD
2
+CD
2
=BE
2
+EC
2
+2ED
2
+2BE.ED−2EC.ED
=BE
2
+EC
2
+2ED
2
+2ED(BE−EC)
=2BE
2
+2ED
2
....[∵BE=EC]
In ΔAED,
AE
2
=ED
2
+AD
2
ED
2
=AE
2
−AD
2
BD
2
+CD
2
=2BE
2
+2AE
2
−2AD
2
.........................(5)
From (5) +(8),
AB
2
+AC
2
=2AD
2
+2BE
2
+2AE
2
−2AD
2
=2BE
2
+2AE
2
Hence proved.
In ΔABD,
AB
2
=BD
2
+AD
2
--(1)
In ΔADC,
AC
2
=CD
2
+AD
2
--(2)
Adding (1) and (2),
AB
2
+AC
2
=2AD
2
+BD
2
+CD
2
...........(8)
BD
2
=BE
2
+ED
2
+2BE.ED --(3)
CD=EC−ED
CD
2
=EC
2
+ED
2
−2EC.ED --(4)
Adding (3) and (4),
BD
2
+CD
2
=BE
2
+EC
2
+2ED
2
+2BE.ED−2EC.ED
=BE
2
+EC
2
+2ED
2
+2ED(BE−EC)
=2BE
2
+2ED
2
....[∵BE=EC]
In ΔAED,
AE
2
=ED
2
+AD
2
ED
2
=AE
2
−AD
2
BD
2
+CD
2
=2BE
2
+2AE
2
−2AD
2
.........................(5)
From (5) +(8),
AB
2
+AC
2
=2AD
2
+2BE
2
+2AE
2
−2AD
2
=2BE
2
+2AE
2
Hence proved
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