Question

State true or false

In a triangle ABC, P and Q are the id-points of AC and AB respectively and angle $BAC={90}^o$.
then $B{P}^2+C{Q}^2=5P{Q}^2$

• A. True
• B. False

Answer 1

The correct option is A True

In $△BAP$,
$B{P}^2=A{B}^2+A{P}^2$ (Pythagoras Theorem)
Since, Q is the mid-point of AB, then $AB=2AQ$
Hence, $B{P}^2=(2AQ{)}^2+A{P}^2$
$B{P}^2=4A{Q}^2+A{P}^2$...(I)
In $triangleQAC$,
$Q{C}^2=A{Q}^2+A{C}^2$ (Pythagoras theorem)
Since, P is the mid point of AC, $AC=2AP$
Hence, $Q{C}^2=A{Q}^2+(2AP{)}^2$
$Q{C}^2=A{Q}^2+4A{P}^2$...(II)
By Adding I and II
$B{P}^2+Q{C}^2=4A{Q}^2+A{P}^2+A{Q}^2+4A{P}^2$
$B{P}^2+Q{C}^2=5(A{P}^2+A{Q}^2)$
In $△QAP$,
$A{Q}^2+A{P}^2=P{Q}^2$
Hence, $B{P}^2+Q{C}^2=5P{Q}^2$