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If $$ XT \perp QR$$ and $$XT \perp PQ $$ prove that :

$$ PX$$ bisects $$\angle P $$

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Solution

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$$ XT \perp QR$$ and $$XT \perp PQ $$

Construction : Draw $$ XZ \cong PR $$ and join $$PX$$

Proof

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

The corresponding parts of the congruent$$ \angle TQX = \angle SQX $$ [$$QX$$ 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,

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

$$XT = XS $$ [ 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 $$\angle P$$

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