Class 9 Mathematics Triangles Textbook - Triangles

Medium

In $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.

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(N/A) Given: $AD$ is the perpendicular bisector of $BC$,which implies $BD = CD$ and $\angle ADB = \angle ADC = 90^o$.
In $\Delta ABD$ and $\Delta ACD$:
$AD = AD$ (Common side)
$\angle ADB = \angle ADC = 90^o$ (Given)
$BD = CD$ (Since $AD$ is the bisector of $BC$)
Therefore,$\Delta ABD \cong \Delta ACD$ by the $SAS$ congruence criterion.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Thus,$AB = AC$.
Since two sides of $\Delta ABC$ are equal,it is an isosceles triangle.

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Textbook - Triangles

In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.

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In an isosceles triangle $ABC$,with $AB = AC$,the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that :
$(i)$ $OB = OC$
$(ii)$ $AO$ bisects $\angle A$

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$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.

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$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see Fig). Show that $\angle A > \angle C$ and $\angle B > \angle D$.

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$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$

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In $\Delta ABC$,$AD$ is the perpendicular bisector of $BC$ (see figure). Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.

Similar Questions

In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.

In an isosceles triangle $ABC$,with $AB = AC$,the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that :$(i)$ $OB = OC$$(ii)$ $AO$ bisects $\angle A$

$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.

$AB$ and $CD$ are respectively the smallest and longest sides of a quadrilateral $ABCD$ (see Fig). Show that $\angle A > \angle C$ and $\angle B > \angle D$.

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:$(i)$ $\Delta DAP \cong \Delta EBP$$(ii)$ $AD = BE$

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In an isosceles triangle $ABC$,with $AB = AC$,the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that :
$(i)$ $OB = OC$
$(ii)$ $AO$ bisects $\angle A$

$AB$ is a line segment and $P$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $\angle BAD = \angle ABE$ and $\angle EPA = \angle DPB$. Show that:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$