$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see Fig). Show that $\angle A > \angle C$ and $\angle B > \angle D$.
(N/A) Let us join $AC$.
In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB < BC$ and $AB < AC$.
Specifically,$BC > AB$.
Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$
In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.
Specifically,$CD > AD$.
Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$
Adding $(1)$ and $(2)$,we get:
$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$
$\Rightarrow \angle A > \angle C$
Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.
In $\Delta ABC$,since $AB$ is the smallest side of the quadrilateral $ABCD$,we have $AB < BC$ and $AB < AC$.
Specifically,$BC > AB$.
Since the angle opposite to the longer side is larger,we have $\angle BAC > \angle BCA$ ........... $(1)$
In $\Delta ACD$,since $CD$ is the longest side of the quadrilateral $ABCD$,we have $CD > AD$ and $CD > AC$.
Specifically,$CD > AD$.
Since the angle opposite to the longer side is larger,we have $\angle CAD > \angle ACD$ ........... $(2)$
Adding $(1)$ and $(2)$,we get:
$(\angle BAC + \angle CAD) > (\angle BCA + \angle ACD)$
$\Rightarrow \angle A > \angle C$
Similarly,by joining $BD$ and using the same logic for $\Delta ABD$ and $\Delta BCD$,we can show that $\angle B > \angle D$.
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Similar Questions
$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.
Medium
View Solution$\Delta ABC$ 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 SolutionIn right triangle $ABC$,right-angled at $C$,$M$ is the mid-point of hypotenuse $AB$. $C$ is joined to $M$ and produced to a point $D$ such that $DM = CM$. Point $D$ is joined to point $B$ (see Fig). Show that:
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
$(i)$ $\Delta AMC \cong \Delta BMD$
$(ii)$ $\angle DBC$ is a right angle.
$(iii)$ $\Delta DBC \cong \Delta ACB$
$(iv)$ $CM = \frac{1}{2} AB$
Difficult
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 SolutionIn the figure,sides $AB$ and $AC$ of $\Delta ABC$ are extended to points $P$ and $Q$ respectively. Also,$\angle PBC < \angle QCB$. Show that $AC > AB$.
Medium
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