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:

1. ΔXTQ ≅ ΔXSQ.
2. 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:

1. ΔXTQ ≅ ΔXSQ.
2. PX bisects angle P.

Construction: Draw XZ ⊥ PR and join PX.

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 and 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.