In triangle ABC obtuse at angle B. AD is perpendicular to BC. Prove AC^2 = AB^2 + BC^2 + 2 BC × BD.
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Given: ABC is obtuse triangle.
AB
2
=AD
2
+DB
2
⟶(1)
AC
2
=AD
2
+DC
2
⟶(2)
from (1) AD
2
=AB
2
−DB
2
∴AC
2
=AB
2
−DB
2
+(DB+BC)
2
[∵BC=DB+BC]
AC
2
=AB
2
−DB
2
+DB
2
+BC
2
+2BC×DB
AC
2
=AB
2
+BC
2
+2BC×DB (Proved)
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