In figure, AD is a median of a triangle ABC and AM is perpendicular to BC. If AB = 4 cm, BC = 6 cm and AC = 8 cm then
find AD.
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Answer:
It is given that
∠AMD=90
0
Referring to the figure, we can say that
∠ADM<90
0
and ∠ADC>90
0
Now,
(i)
To prove:
AC
2
=AD
2
+BC.DM+(
2
BC
)
2
In ΔADC,∠ADC ia an obtuse angle.
∴AC
2
=Ad
2
+DC
2
+2DC.DM
⇒AC
2
=AD
2
+(
2
BC
)
2
+2.
2
BC
.DM
⇒AC
2
=AD
2
+BC.DM+(
2
BC
)
2
(ii)
To prove:
AB
2
=AD
2
−BC.DM+(
2
BC
)
2
In ΔABD,∠ADM is an obtuse angle.
∴AB
2
=AD
2
+BD
2
−2BD.DM
⇒AB
2
=AD
2
+(
2
BC
)
2
−2.
2
BC
.DM
⇒AB
2
=AD
2
−BC.DM+(
2
BC
)
2
(iii)
To prove:
AC
2
+AB
2
=2AD
2
+
2
1
BC
2
From the result of (i) and (ii), adding those, we get
AC
2
+AB
2
=2AD
2
+
2
1
BC
2
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