in the given figure, AD PERPENDICULAR BC and BE = EC IF DC = 3/4 BC and DE =1/4 BC then prove that BC² = 2AC² - 2AE²
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Answer:
In ΔABC, seg AD ⊥ seg BC , DB =3CD. Prove that: 2AB
2
=2AC
2
+BC
2
.
Step-by-step explanation:
Solution
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We are given
here
AD
ˉ
⊥
BC
ˉ
& DB=3CD−−−(1)
so, BC=DB+CD
BC=3CD+CD
so, BC=4CD−−−(2)
Hence it given that,
AB
ˉ
is perpendicular to
BC
ˉ
by the pythagoras theorem
In both triangle △ABD & △ACD
AB
2
=AD
2
+BD
2
& AC
2
=AD
2
+CD
2
∴ AD
2
=AB
2
−DB
2
& AD
2
=AC
2
−CD
2
Compare both AD
2
with each other we get,,
AB
2
−DB
2
=AC
2
−CD
2
AB
2
=AC
2
+DB
2
−CD
2
AB
2
=AC
2
+(3 CD)
2
−CD
2
(from (1))
AB
2
=AC
2
+9 CD
2
−CD
2
AB
2
=AC
2
+8 CD
2
AB
2
=AC
2
+8(
4
BC
)
2
(from (2))
2AB
2
=2 AC
2
+BC
2
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