Given in BCD, BD = BC and BE is parallel to CD. [3] Prove that 1=2.
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Answer:
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
(Let multiply equation with 2)
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