Question

In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR.
If XS ⊥ QR and XT ⊥  PQ ;

prove that: (i) ΔXTQ ≅ ΔXSQ. 
                   (ii) PX bisects angle P.

Answer 1

Given: A( ΔPQR ) in which QX is the bisector of ∠Q. and RX is the bisector of ∠R.
XS ⊥ QR and XT ⊥  PQ.

We need to prove that

(i) ΔXTQ ≅ ΔXSQ.

(ii) PX bisects angle P.

Construction: Draw XZ ⊥ PR and join PX.

Proof: 
(i) In ΔXTQ and ΔXSQ,

∠QTX = ∠QSX = 90°    ...[ XS ⊥ QR and XT ⊥  PQ ]

∠TQX = ∠SQX              ...[ QX is bisector of ∠Q ]

QX = QX                      ...[ Common ]

∴ By Angle-Side-Angle Criterion of congruence,

ΔXTQ ≅ ΔXSQ

(ii) The corresponding parts of the congruent triangles are congruent.

∴ XT = XS           ...[ c.p.c.t. ]

In ΔXSR & ΔXRZ

∠XSR = ∠XZR = 90°   ...[ XS ⊥ QR and ∠XSR = 90° ]

∠XRS = ∠ZRX             ...[ RX is bisector of ∠R ]

RX = RX                      ....[ Common ]

∴ By Angle-Angle-Side criterion of congruence,

ΔXSR ≅ ΔXRZ

The corresponding parts of the congruent triangles are congruent.

∴ XS = XT             ...[ c.p.c.t. ] 

From (1) and (2)

XT = XZ                    

In ΔXTP and ΔPZX

∠XTP = ∠XZP = 90°       ....[ Given ]

XP = XP         ....[ Common ]

XT = XZ               

∴ By Right angle-Hypotenuse-side criterion of congruence,

ΔXTP ≅ ΔPZX

The corresponding parts of the congruent triangles are
congruent.

∴ ∠TPX = ∠ZPX          ...[ c.p.c.t. ]

∴ PX bisects ∠P.