Question

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 ​​

Answer 1

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