Let PQR be am acute-angled triangle in which PQ lt QR. From the vertex Q draw altitude `QQ_(1)`, the angle bisector `Q Q_(2)` and the median`Q Q_(3)" With "Q_(1),Q_(2),Q_(3)` lying on PR. Then

Options

A. `PQ_(1)ltPQ_(2)ltPQ_(3)`
B. `PQ_(2)ltPQ_(1)ltPQ_(3)`
C. `PQ_(1)ltPQ_(3)ltPQ_(2)`
D. `PQ_(3)ltPQ_(1)ltPQ_(2)`

Answer

A

Explanation

`PQ_(3)=Q_(3)R( :. Q Q_(3)" ie median")`
`PQ_(3)=1/2PR`
`PQ_(2):Q_(2)R=r:p" (By peoperty of angle bisector)"`
`PQ_(2)=(r/(r+P))PR`
But r lt P (Given)
`PQ_(2)lt1/2PR`
Comparison between Altitude and angle bisector
`rArr angleQPQ_(2)+anglePQ_(2)Q+anglePQ Q_(2)=angle RQ Q_(2)+angleQ Q_(2)R+angleQRQ_(2)`
`:. anglePQ Q_(2)+angleRQ Q_(2)" {Since angle bisector}"`
`angle QPQ_(2)+anglePQ_(2)Q=angle Q Q_(2)R+angleQRQ_(2)`
`:.PQltQR" then"ltQPQ_(2)gtangle QRQ_(2)`
Hence `angleQ Q_(2)P+ltangleQ Q_(2)R`
But `angle Q Q_(2)P+angleQ Q_(2)R=180^(@)`
Hence `angle Q Q_(32)Plt90^(@)&angle Q Q_(2)Rgt 90^(@)`
`rArr" Foot from Q to side PR lies inside "DeltaPQ Q_(2)`
`rArrPQ_(1)lt PQ_(2)ltPQ_(3)`