In triangle ABC , angle C=90^ , A - D - E - B . AD = DE = EB . show that, 2AB =3(CD^2 + CE^2 +DE^2).
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Answer:
In △ABC,∠A=90
AC=AB
D is an AB product
To prove :DC
2
−BD
2
=
2AB×AD
→ Acc to Pythagoras theorem
In △ACD
CD
2
=AC
2
+AD
.
..(1)
and In △ABC(∵AB+BD=AD)
BC
2
=AC
2
+AB
2
...(2)
from (1):
CD
2
=AC
2
+AD
2
CD
2
−BD
2
=AC
2
+AD
2
−BD
2
CD
2
−BD
2
=AC
2
+AD
2
−(AD−AB)
2
=AC
2
+AD
2
−AD
2
−AB
2
+2(AD/AB)
CD
2
−BD
2
=AC
2
+(−AB
2
)+2(AD)(AB)
=AB
2
−AB
2
+2(AB)(AD)
or (∵AB=AC)
CD
2
=BD
2
=2AB×AD
Hence proved
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