Question

plz. answer this fast

Answer 1

in ∆BPC,

$bp {}^{2} =bc {}^{2} + pc {}^{2}$
in∆ AQC,
$aq {}^{2} = ac {}^{2} + qc {}^{2}$
add 1st and 2nd equation

$bp {}^{2} + aq {}^{2} = bc {}^{2} + ac {}^{2} + pc {}^{2} + qc {}^{2}$
bp^2 +aq^2=ab^2+ (1/2 AC)^2 +(1/2BC)^2

bp^2 + aq^2=ab^2 +1/4AC^2 +1/4BC^2

bp^2+ aq^2= (4ab^2 +ab^2 )/4

4(bp^2 +aq^2)=5ab^2

hence proved

Answer 2

REFER TO THE ATTACHMENT..

HOPE IT HELPS..!!! ☺☺