Question

In triangle BAC,angle BCA is a right angle and Q is the midpt of BC P.T. BC^2=4(AQ^2-AC^2).
Pls give step by step solution. It's urgent. I will surely mark u as brainliest answer.​

Answer 1

Answer:

AQ

2

=AC

2

+(

3

2

BC)

2

2)9AQ

2

=9AC

2

+4BC

2

−−−−−(A)

In BCP

CA−CP

CP

=2

PB

2

=BC

2

+PC

2

−−CP=

3

2

CA

PB

2

=BC

2

+(

3

2

CA)

2

⇒9PB

2

=9BC

2

+4AC

2

−−−−−(B)

(3) Adding (A)&(B)

9AQ

2

=9BC

2

=13(BC

2

+AC

2

)

⇒9(AQ

2

+BP

2

)=13AB

2