Class 9 Mathematics Triangles Textbook - Triangles

Medium

In the figure,$OA = OB$ and $OD = OC$. Show that
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.

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Solution

(N/A) $(i)$ In $\Delta AOD$ and $\Delta BOC$:
$OA = OB$ (Given)
$OD = OC$ (Given)
Also,$\angle AOD = \angle BOC$ (Vertically opposite angles).
Therefore,by the $SAS$ congruence rule,$\Delta AOD \cong \Delta BOC$.
$(ii)$ Since $\Delta AOD \cong \Delta BOC$,their corresponding parts are equal $(CPCT)$.
Thus,$\angle OAD = \angle OBC$.
These are alternate interior angles formed by the transversal $AB$ intersecting lines $AD$ and $BC$.
Since the alternate interior angles are equal,$AD \parallel BC$.

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Textbook - Triangles

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In the figure,$OA = OB$ and $OD = OC$. Show that$(i)$ $\Delta AOD \cong \Delta BOC$ and$(ii)$ $AD \parallel BC$.

Similar Questions

In $\Delta ABC$,the bisector $AD$ of $\angle A$ is perpendicular to side $BC$ (see figure). Show that $AB = AC$ and $\Delta ABC$ is isosceles.

Line $l$ 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 $\angle A$.

In the figure,sides $AB$ and $AC$ of $\Delta ABC$ are extended to points $P$ and $Q$ respectively. Also,$\angle PBC < \angle QCB$. Show that $AC > AB$.

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In the figure,$OA = OB$ and $OD = OC$. Show that
$(i)$ $\Delta AOD \cong \Delta BOC$ and
$(ii)$ $AD \parallel BC$.

Line $l$ 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 $\angle A$.