Question

In atriangle ABC angle C=90 and p and q are the points on BC and CA respectively such that CP =1/3BC and CQ =1/3AC Prove that 9(AP^2+BQ^2)=10AB^2

Answer 1

Given it

CQ=1/3 AC

PC=1/3 BC

To proof that

9 (AP^2+BQ^2)=10AB^2

In ABC

AB^2=AC^2+BC^2

In APC

AP^2=AC^2+PC^2

AP^2=AC^2+(1/3BC)^2

AP^2=AC^2+(BC^2/9)

9AP^2=9AC^2+BC^2......... (1)

In BCQ

BQ^2=CQ^2+BC^2

BQ^2=(1/3AC)^2+BC^2

BQ^2=(AC^2/9)+BC^2

9BQ^2=9BC^2+AC^2....... (2)

Sum of eq (1)&(2)

9AP^2+9BQ^2=9AC^2+BC^2+9BC^2+ AC^2

9 (AP^2+BQ^2)=10AC^2+10BC^2

9 (AP^2+BQ^2)=10 (AC^2+BC^2)

9 (AP^2+BQ^2)=10 BC^2