Question

In a triangle QX,RX are bisectors of angle PQR and PRQ respectively of triangle PQR.If XS perpendicular to QR and XT perpendicular to PQ,prove that triangle XTQ is congruent to triangle XSQ​

Answer 1

Answer:

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

Answer 2

Answer:

Hello

Step-by-step explanation:

$\huge\bold\pink{Answer}$

As the figure is not given, but I have tried my best to solve this question.

Solution:

Angle XSQ = 90°

Angle XTQ = 90°

QX = QX (This is the common side)

Since QX bisects Q the angle is equally split between the triangles)

So, XQS = TQX

So, Δ XTQ is congruent to Δ XSQ

XS = XT (According to CPCT) ...(1)

Draw XW perpendicular to PR

Similarly, we can prove that Δ XSR is congruent to Δ XWR.

So, XS = XW ... (2)

So, from (1) and (2)

now in PXT and PXW, PTX = PWX, PX is common and XT = XW.

BY R.H.S. both triangles are congruent, XPT = XPW and PX bisects the angle P. Hence proved.

Hope it's help uh...!!!