In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR. If XS is perpendicular to QR and XT is perpendicular to PQ. Prove that triangle XTQ is congruent to triangle XSQ and PX bisect angle P.

We have

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.