Question

In the adjoining figure ,QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR .
If $$ XT \perp QR$$ and $$XT \perp PQ $$ prove that :
$$ PX$$ bisects $$\angle P $$

Answer 1

Given : A $$ \Delta PQR $$ in which $$QX$$ is the bisectors of $$\angle Q $$ and $$RX$$ is the bisectors of $$\angle R $$
$$ XT \perp QR$$ and $$XT \perp PQ $$
Construction : Draw $$ XZ \cong PR $$ and join $$PX$$
Proof

In $$ \Delta XTQ $$ and $$\Delta XSQ $$

$$ \angle TQX = \angle SQX $$ [$$QX$$ is the angle bisector of $$\angle Q$$]

$$ \angle XTQ = \angle XSQ=90^0 $$ [Perpendicular to sides]

$$QX = QX$$ [Common]

By Angle - Angle - Side criterion of congruence,

$$ \Delta XTQ \cong \Delta XSQ $$

The corresponding parts of the congruent
$$XT = XS $$ [ c.p.c.t]
In $$\Delta XSR$$ and $$\Delta XZR$$
$$\angle XSR=\angle XZR=90^0$$ ...[$$XS \bot QR$$ and $$\angle XSR=90^0$$]
$$\angle SRX=\angle ZRX ...[$$RX$$ is bisector of $$\angle R$$]
$$RX=RX ...[Common]
By Angle-Angle-Side criterion of congruence,
$$\Delta XSR\cong \Delta XZR$$
The corresponding parts of the congruent triangles are equal.
$$\therefore XS=XZ$$ ...[C.P.C.T] ...(2)
From (1) and (2)
$$XT=XZ$$ ...(3)
In $$\Delta XTP$$ and $$\Delta XZP$$
$$\angle XTP=\angle XZP=90^0$$ ...[Given]
$$XP=XP$$ ...[Common]
$$XT=XZ$$ ...[From (3)]
The corresponding parts of the congruent triangles are congruent

$$ \angle XPT = \angle XPZ $$

So, $$PX$$ bisects $$\angle P$$