REPOSTING THE QUESTION
In the given figure, QX and RX are the bisectors of ∠Q and ∠R respectively of the △PQR. If XS ⊥ QR and XT⊥ PQ, prove that PX bisects the ∠P
Answers
Step-by-step explanation:
Given : A ΔPQR in which QX is the bisectors of ∠Q and RX is the bisectors of ∠R
XT⊥QR and XT⊥PQ
Construction : Draw XZ≅PR and join PX
Proof
In ΔXTQ and ΔXSQ
∠TQX=∠SQX [QX is the angle bisector of ∠Q]
∠XTQ=∠XSQ=90
0
[Perpendicular to sides]
QX=QX [Common]
By Angle - Angle - Side criterion of congruence,
ΔXTQ≅ΔXSQ
The corresponding parts of the congruent
XT=XS [ c.p.c.t]
In ΔXSR and ΔXZR
∠XSR=∠XZR=90
0
...[XS⊥QR and ∠XSR=90
0
]
∠SRX=∠ZRX...[RXisbisectorof\angle R]RX=RX ...[Common]
By Angle-Angle-Side criterion of congruence,
ΔXSR≅ΔXZR
The corresponding parts of the congruent triangles are equal.
∴XS=XZ ...[C.P.C.T] ...(2)
From (1) and (2)
XT=XZ ...(3)
In ΔXTP and ΔXZP
∠XTP=∠XZP=90
0
...[Given]
XP=XP ...[Common]
XT=XZ ...[From (3)]
The corresponding parts of the congruent triangles are congruent
∠XPT=∠XPZ
So, PX bisects ∠P