Question

In ` P Q R , Q M_|_P R` and `P R^2-P Q^2=Q R^2` . Prove that `Q M^2=P MxxM R`

Answer 1

since ` PR^(2)-PQ^(2)= QR^(2)`
` Rightarrow " " PR^(2)= PQ^(2) + QR^(2)`
` Rightarrow " " angle4+ angle5 = 90^(@)` ( by the converse of pythagoras theorem)
Now, since , ` angle2 = 90^(@)`
` Rightarrow " " angle3 + angle4 = 90^(@)` (angle sum property) ...(1)
But `angle4 + angle5 = 90^(@)`
From (1) and (2) ,we get
`angle3 + angle4 = angle4 + angle5`
`Rightarrow " " angle3 = angle5`
Now in `triangle PMQ and triangleQMR`
( corresponding sides of similar triangle are proportional)
` Rightarrow " " `QM^(2) = PM xx MR`