In Figure `P S\ ` is the bisector of `/_Q P R\ a n d\ P T\ _|_Q R` . Show that `/_T P S=1/2(/_Q-/_R)`

Explanation

Given In `Delta PQR, /_gt /_R`, PA is the bisector of `/_QPR and PM bot` QR .
To prove `/_APM = 1/2( /_Q- /_R)`
Proof Since, PA is the bisector of `/_QPR`
`:." " /_QPA = /_APR " "`....(i)
In `DeltaPQM, " "/_ PQM +/_PMQ + /_QPM +180^(@) " "`[by angle sum property of a triangle]
`rArr " "/_PQM + 90^(@) + /_QPM = 180^(@)" "[:'PMbotQRrArr/_PMQ = 90^(@)`]
`rArr " "/_PQM + 90^(@) - /_QPM " "...(ii) `
In `DeltaPMR, " "/_PMR + /_PRM + /_ RPM =180^(@) " "`[by angles sum property of a triangles]
`rArr " "/_90^(@) + /_PRM + /_ RPM =180^(@) " "[:'PMbotQRrArr/_PMQ = 90^(@)`]
`rArr " " /_PRM = 180^(@)-90^(@)-/_RPM`
`rArr " " /_PRM = 90^(@)-/_RPM " "` ...(iii)
On subtracting Eq. (iii) from Eq. (ii), we get
`/_Q - /_R = (90^(@)-/_QPM)-(90^(@)-/_RPM) " " [where/_PQM = /_Q and /_PRM = /_R]`
`rArr" "/_Q -/_R = /_RPM - /_QPM`
`rArr" "/_Q -/_R = [/_APM + /_APM]-[/_QPA - /_APM] " "`...(iv)
`rArr" "/_Q -/_R = /_QPA + /_APM-/_QPA - /_APM" " `[by using Eq. (i)]
`rArr" "/_Q -/_R = 2/_APM`
`:." " /_APM = 1/2(/_Q - /_R) " "` Hence proved.