In the adjoining figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR. If XS ⊥ QR and XT ⊥ PQ; prove that :

(i) △ XTQ ≅ △ XSQ

(ii) PX bisects angle P.

(i) In △ XTQ and △ XSQ,

⇒ ∠XSQ = ∠XTQ (Both equal to 90°)

⇒ XQ = XQ (Common side)

⇒ ∠XQT = ∠XQS (Since, XQ is the bisector of ∠Q)

∴ △ XTQ ≅ △ XSQ (By A.A.S. 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 of ∠R)

⇒ XR = XR (Common side)

∴ △ XSR ≅ △ XUR (By A.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ XU = XS ……(1)

As, △ XTQ ≅ △ XSQ

∴ XS = XT ……(2)

In △ XUP and △ XTP,

From (1) and (2) we get,

⇒ XU = XT

⇒ XP = XP (Common)

⇒ ∠XTP = ∠XUP (Both are equal to 90°)

∴ △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.