In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of △PQR. If XS ⊥ QR and XT ⊥ PQ, prove that

(i) △XTQ ≅ △XSQ

(ii) PX bisects the angle P.

(i) Given, QX is the bisector of ∠PQR

∴ ∠PQX = ∠XQS

From figure,

∠XTQ = ∠XSQ (Both are equal to 90°)

XQ = XQ

Hence, △XTQ ≅ △XSQ by ASA axiom.

Hence, proved that △XTQ ≅ △XSQ.

(ii) Draw a perpendicular from X on PR i.e. XU.

In △XSR and △XUR,

∠XSR = ∠XUR (Both are equal to 90°)

∠XRS = ∠XRU (As XR is bisector)

XR = XR (Common)

Hence, △XSR ≅ △XUR by AAS axiom.

We know that corresponding parts of congruent triangle are equal.

∴ XU = XS ……..(i)

As, △XTQ ≅ △XSQ

∴ XS = XT …….(ii)

In △XUP and △XTP,

From (i) and (ii) we get,

XU = XT

XP = XP (Common)

∠XTP = ∠XUP (Both are equal to 90°)

Hence, △XUP ≅ △XTP by SAS axiom.

We know that corresponding parts of congruent triangle are equal.

∴ ∠XPU = ∠XPT

Hence, proved that PX is bisector of ∠P.