Question

please answer this simple proof question.please.....

Answer 1

=_= 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

Answer 2

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.