Class 9 Mathematics Triangles Textbook - Triangles

Medium

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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Solution

(N/A) In $\Delta ABD$ and $\Delta ACE$:
$AB = AC$ (Given) ........... $(1)$
$\angle B = \angle C$ (Angles opposite to equal sides) ........... $(2)$
Also,$BE = CD$
Subtracting $DE$ from both sides:
$BE - DE = CD - DE$
$BD = CE$ ........... $(3)$
Therefore,$\Delta ABD \cong \Delta ACE$ (Using $(1)$,$(2)$,$(3)$ and $SAS$ congruence rule).
This gives $AD = AE$ $(CPCT)$.

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Triangles

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Textbook - Triangles

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

Medium

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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 $\angle A$.

Medium

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Show that in a right-angled triangle,the hypotenuse is the longest side.

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$\Delta ABC$ 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.

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

Similar Questions

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

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 $\angle A$.

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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 $\angle A$.