Question

P&Q respectively are midpoints of the sides CA;CB of a right angle triangle ABC right angled at C prove that
1) $4A{Q}^2=4A{C}^2+B{C}^2$
2) $4B{P}^2=4B{C}^2+A{C}^2$
3) $4(A{Q}^2+B{P}^2)=5A{B}^2$

Answer 1

Given P and Q are points of the sides of the sides CA and CB respectively, which divides these triangles in the ratio of 2 : 1. So,
$CQ=\frac{2}{2+1}BC=\frac{2}{3}BC-------\left(A\right)AndCP=\frac{2}{2+1}=AC=\frac{2}{3}AC----------\left(B\right)$

i) Apply Pythagoras theorem in $\Delta ACQ,$ and get
$A{Q}^2=A{C}^2+C{Q}^2$, Substitute value from equation (A) and get
$A{Q}^2=A{C}^2+(\frac{2}{3}BC{)}^2$
$A{Q}^2=A{C}^2+\frac{4}{9}B{C}^2$
Taking LCM we get
$9A{Q}^2=9A{C}^2+4B{C}^2--------\left(1\right)$ (Hence proved)
(ii) Apply Pythagoras theorem in $\Delta PCB$, and get
$B{P}^2=C{P}^2+B{C}^2,$ Substitute value from equation (B) and get
$B{P}^2=B{C}^2+(\frac{2}{3}AC{)}^{2}B{P}^2=B{C}^2+\frac{4}{9}A{C}^2$
Taking LCM we get
$9B{P}^2=9B{C}^2+4A{C}^2----------\left(2\right)$ (Hence proved)
iii) Apply Pythagoras theorem in $\Delta ABC$, and get
$A{B}^2=A{C}^2+B{C}^2-----------\left(3\right)$
Now we add equation 2 and 3, we get
$9A{Q}^2+9B{P}^2=9A{C}^2+4B{C}^2+9B{C}^2+4A{C}^2$
$9(A{Q}^2+B{P}^2)=13A{C}^2+13B{C}^2$
$9(A{Q}^2+B{P}^2)=13\left(A{B}^2\right)$ ( Hence proved)
2) Given P and Q are midpoints of the sides CA and CB respectively.
So,
$CQ=QB=\frac{1}{2}BC-------\left(A\right)$
And
$CP=PA=\frac{1}{2}AC-------\left(B\right)$

i) Apply Pythagoras theorem in $\Delta ACQ,$ and get
$A{Q}^2=A{C}^2+C{Q}^2$, Substitute value from equation (A) and get
$AQ2=AC2+(\frac{1}{2}BC{)}^2$
$A{Q}^2=A{C}^2+\frac{1}{4}B{C}^2$
Taking LCM we get
$4A{Q}^2=4A{C}^2+B{C}^2-------\left(1\right)$
Now, Apply Pythagoras theorem in \Delta PCB, and get\\
$B{P}^2=C{P}^2+B{C}^2,$ Substitute value from equation (B) and get\\
$B{P}^2=B{C}^2+(\frac{1}{2}AC{)}^{2}B{P}^2=B{C}^2+\frac{1}{4}A{C}^2$
Taking LCM we get
$4B{P}^2=4B{C}^2+A{C}^2---------\left(2\right)$
And, Apply Pythagoras theorem in \Delta ABC, and get\\
$A{B}^2=A{C}^2+B{C}^2---------\left(3\right)$
Now we add equation 2 and 3, we get
$4A{Q}^2+4B{P}^2=4A{C}^2+B{C}^2+4B{C}^2+A{C}^2$
$4(A{Q}^2+B{P}^2)=5A{C}^2+5B{C}^2$
$4(A{Q}^2+B{P}^2)=5(A{C}^2+B{C}^2)$, Now substitute value from equation 3 and get
$4(A{Q}^2+B{P}^2)=5\left(A{B}^2\right)$ (Hence proved)