in equilateral triangle ABC if AD perpendicular to BC then prove 3AB2=4AD2
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Answered by
17
In ADB,
AB²=AD²+BD²
AB²-BD²=AD²
AB²-(BC/2)²=AD²
AB²-BC²/4=AD²
(4AB²-BC²)/4=AD²
4AB²-AB²=4AD²(because AB=BC)
3AB²=4AD2
AB²=AD²+BD²
AB²-BD²=AD²
AB²-(BC/2)²=AD²
AB²-BC²/4=AD²
(4AB²-BC²)/4=AD²
4AB²-AB²=4AD²(because AB=BC)
3AB²=4AD2
Answered by
8
in triangle ABC,
sides are 'a' units
as AD is perpendicular to BC ,
then in triangle ABD ,
AB2 = AD2 + BD2
a2 = AD2 + (a/2)2 [ since AD is perpendicular to BC ]
a2 - a2/4 =AD2
= 3a2 /4 = AD2 ..........(1)
in triangle ADC,
AC2 = AD2 + CD2
a2 = AD2 + (a/4)2
3a2/4 = AD2 .........(2)
from equation 1 and 2
3 a2 /4 = AD2
3 a2 = 4 AD2
3 AB2 = 4AD2 [since AB = a ]
hence proved
sides are 'a' units
as AD is perpendicular to BC ,
then in triangle ABD ,
AB2 = AD2 + BD2
a2 = AD2 + (a/2)2 [ since AD is perpendicular to BC ]
a2 - a2/4 =AD2
= 3a2 /4 = AD2 ..........(1)
in triangle ADC,
AC2 = AD2 + CD2
a2 = AD2 + (a/4)2
3a2/4 = AD2 .........(2)
from equation 1 and 2
3 a2 /4 = AD2
3 a2 = 4 AD2
3 AB2 = 4AD2 [since AB = a ]
hence proved
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