Question

12. In the given figure, QX and RX are the bisectors
of angle Q and angle R respectively of ∆PQR. If XS perpendicular QR
and XT 1 PQ, prove that
(i) ∆XTQ = ∆XSQ
(ii) PX bisects angle P

If any one will give the answer then I will mark him/her as a brainlist . please help


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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

Step-by-step explanation:

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Answer 2

Answer:

Given: In ∆PQR,

QX bisects the ∠Q & RX bisects the ∠R.

XS ⊥ QR & XT ⊥ PQ.

To Prove: ∆XTQ ≅ ∆XSQ

& PX bisects ∠P.

Proof: In ∆XSQ and ∆XTQ, we have

∠XSQ = 90°

∠XTQ = 90°

⇒ ∠XSQ = ∠XTQ …(i)

QX = QX [∵ This is the common side] …(ii)

∠XQS = ∠TQX [∵ it is given that QX bisects ∠Q] …(iii)

So, by (i), (ii) & (iii), we can say that

ΔXTQ ≅ ΔXSQ by AAS rule of congruency.

⇒ XS = XT [∵ corresponding parts of congruent triangles are always equal] …(iv)

Draw XW perpendicular to PR.

We have

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

So, XS = XW …(v)

So, from (iv) and (v), we can say

XT = XW …(vi)

In ∆PXT and ∆PXW,

∠PTX = ∠PWX [∵ ∠PTX = ∠PWX = 90°]

PX = PX [∵ they are common sides]

XT = XW [from (vi)]

BY R.H.S. both triangles are congruent,

That is, ∆PXT ≅ ∆PXW.

⇒ ∠XPT = ∠XPW [∵ corresponding parts of congruent triangles are always equal]

⇒ PX is bisector of ∠P.

Hence, proved.