Question

please solve I want the solution ​

Answer 1

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.

Step-by-step explanation: