please answer this simple proof question.please.....
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please answer this simple proof question.please.....
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=_= A lengthy question--> I would avoid the stupid unnecessary lines :p
( i ) --> 4b² = 4 ( h² + EC² )
= 4h² + 4 ( a/2 + x )²
= 4h² + ( a + 2x )²
= 4h² + 4x² + 4ax + a²
= 4 ( h² + x² ) + 4ax + a²
= 4p² + 4ax + a²
=> b² = p² + ax + a²/4 { and we are done =_= }
( ii ) --> 4c² = 4 ( h² + BE² )
= 4h² + 4 ( a/2 - x )²
= 4h² + ( a - 2x )²
= 4h² + 4x² - 4ax + a²
= 4 ( h² + x² ) - 4ax + a²
= 4p² - 4ax + a²
=> c² = p² - ax + a²/4 --> { copy pasted the above proof =_= }
( iii ) --> Adding ( i ) and ( ii ) -->
----> b² + c² = ( p² + ax + a²/4 ) + ( p² - ax + a²/4 )
==>b² + c² = 2p² + a²/2
^_^ Now, may I tell you that a little change in ( iii ) is known as APPOLONIUS THEOREM -->
--> ( iii ) => b² + c² = 2p² + a²/2
=> b² + c² = 2 [ p² + (a/2)² ]
=> AB² + AC² = 2 [ AD² + BD² ]
And there, we have all are proofs done ^_^
---> Hope you liked the answer :p
Step-by-step explanation:
∠P=∠Q=90
o
∠PAB=∠QAC [Vertically opposite angles]
∴ △BPA∼△AQC [AAA similarity criterion]
⇒
AC
AB
=
AQ
AP
∴ AB×AQ=AC×AP --- ( 1 ) [ Hence proved ]
Consider, right angled △BCQ
⇒ BC
2
=CQ
2
+BQ
2
[By Pythagoras theorem]
⇒ BC
2
=CQ
2
+(AB+AQ)
2
[Since BQ = AB + AQ]
⇒ BC
2
=[CQ
2
+AQ
2
]+AB
2
+2AB×AQ ----- (2)
In right △ACQ, CQ
2
+AQ
2
=AC
2
[By Pythagoras theorem]
Hence equation (2) becomes,
⇒ BC
2
=AC
2
+AB
2
+AB×AQ+AB×AQ
⇒ BC
2
=AC
2
+AB
2
+AB×AQ+AP×AC [From (1)]
⇒ BC
2
=AC
2
+AP×AC+AB
2
+AB×AQ
⇒ BC
2
=AC(AC+AP)+AB(AB+AQ)
⇒ BC
2
=AC×CP+AB×BQ [From the figure]
Hence Proved.