In the figure,$PR > PQ$ and $PS$ bisects $\angle QPR$. Prove that $\angle PSR > \angle PSQ$.
(N/A) In $\Delta PQR$,$PS$ bisects $\angle QPR$ [Given].
Therefore,$\angle QPS = \angle RPS$.
Since $PR > PQ$ [Given],
Therefore,the angle opposite to $PR$ is greater than the angle opposite to $PQ$.
Thus,$\angle PQS > \angle PRS$.
Adding $\angle QPS$ to the left side and $\angle RPS$ to the right side (since $\angle QPS = \angle RPS$):
$\angle PQS + \angle QPS > \angle PRS + \angle RPS$ ... $(1)$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta PQS$,the exterior angle $\angle PSR = \angle PQS + \angle QPS$.
For $\Delta PRS$,the exterior angle $\angle PSQ = \angle PRS + \angle RPS$.
Substituting these into $(1)$,we get $\angle PSR > \angle PSQ$.
Therefore,$\angle QPS = \angle RPS$.
Since $PR > PQ$ [Given],
Therefore,the angle opposite to $PR$ is greater than the angle opposite to $PQ$.
Thus,$\angle PQS > \angle PRS$.
Adding $\angle QPS$ to the left side and $\angle RPS$ to the right side (since $\angle QPS = \angle RPS$):
$\angle PQS + \angle QPS > \angle PRS + \angle RPS$ ... $(1)$.
We know that the exterior angle of a triangle is equal to the sum of the two interior opposite angles.
For $\Delta PQS$,the exterior angle $\angle PSR = \angle PQS + \angle QPS$.
For $\Delta PRS$,the exterior angle $\angle PSQ = \angle PRS + \angle RPS$.
Substituting these into $(1)$,we get $\angle PSR > \angle PSQ$.
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Similar Questions
$\Delta ABC$ and $\Delta DBC$ are two isosceles triangles on the same base $BC$ and vertices $A$ and $D$ are on the same side of $BC$ (see Fig). If $AD$ is extended to intersect $BC$ at $P$,show that:
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
$(i)$ $\Delta ABD \cong \Delta ACD$
$(ii)$ $\Delta ABP \cong \Delta ACP$
$(iii)$ $AP$ bisects $\angle A$ as well as $\angle D$.
$(iv)$ $AP$ is the perpendicular bisector of $BC$.
Difficult
View Solution$ABC$ is a triangle. Locate a point in the interior of $\Delta ABC$ which is equidistant from all the vertices of $\Delta ABC$.
Medium
View Solution$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.
Medium
View SolutionLine $l$ is the bisector of an angle $\angle A$ and $B$ is any point on $l$. $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $\angle A$ (see Fig). Show that :
$(i)$ $\Delta APB \cong \Delta AQB$
$(ii)$ $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
$(i)$ $\Delta APB \cong \Delta AQB$
$(ii)$ $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
Medium
View Solution$\Delta ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$ (see figure). Show that $\angle BCD$ is a right angle.
Difficult
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