Question

if p and q are the midpoints of the sides ca and cb respectively of a triangle abc in which c is a right angle then 4(aq²+bp²)0=

Answer 1

In ΔACQ
AQ² = AC² + QC²
AQ² = AC² + (BC/2)²
AQ² = AC² + BC²/4    .............(1)

In ΔBPC
BP² = BC² + PC²
BP² = BC² + (AC/2)²
BP² = BC² + AC²/4        ...........(2)

Adding (1) and (2)
AQ² + BP² = AC² + BC² + AC²/4 + BC²/4 

4(AQ² + BP²) = 5AC² + 5BC²  
4(AQ² + BP²) = 5(AC² + BC²)
4(AQ² + BP²) = 5(AB²)
  

Answer 2

Answer:

If p and q are the midpoints of the sides $CA$ and CB respectively of a triangle ABC in which c is a right angle then $4(AQ^2+BP^2)=5AB^2$

Step-by-step explanation:

Given that p and q are the midpoints of the sides $CA$ and CB respectively of a triangle ABC in which c is a right angle.

Since △AQC is a right-angled triangle at C.

By Pythagoras theorem

 $AQ^2=AC^2+QC^2$

Multiplying both sides by $4$

$4AQ^2=4AC^2+4QC^2\\4AQ^2=4AC^2+(2QC)^2\\4AQ^2=4AC^2+BC^2...(1)$

Now since $\Delta BPC$ is a right-angled triangle at $C$.

By Pythagoras theorem

$BP^2=BC^2+CP^2$

Multiplying both sides by $4$

$4 BP^2=4 BC^2+4 CP^2\\4BP^2=4BC^2+(2CP)^2\\4BP^2=4BC^2+AC^2...(2)$

Now from equations (1) and (2) we have

$4AQ^2+4BP^2=(4AC^2+BC^2)+(4BC^2+AC^2)\\4(AQ^2+BP^2)=5(AC^2+BC^2)\\4(AQ^2+BP^2)=5AB^2( \because AB^2=AC^2+BC^2)$