Class 9 Mathematics Triangles Textbook - Triangles

Medium

$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ bisects $\angle A$.

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Solution

(N/A) $(i)$ In $\Delta ABD$ and $\Delta ACD$,we have:
$AB = AC$ [Given]
$\angle ADB = \angle ADC = 90^\circ$ [Since $AD$ is an altitude]
$AD = AD$ [Common side]
Therefore,by $RHS$ congruence rule,$\Delta ABD \cong \Delta ACD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$BD = CD$,which means $AD$ bisects $BC$.
$(ii)$ Since $\Delta ABD \cong \Delta ACD$,their corresponding angles are equal $(CPCT)$.
Therefore,$\angle BAD = \angle CAD$,which means $AD$ bisects $\angle A$.

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

Medium

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$E$ and $F$ are respectively the mid-points of equal sides $AB$ and $AC$ of $\Delta ABC$ (see figure). Show that $BF = CE$.

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$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig). Show that $\Delta ABC \cong \Delta CDA$.

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$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.

Medium

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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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$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:$(i)$ $AD$ bisects $BC$$(ii)$ $AD$ bisects $\angle A$.

Similar Questions

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

$E$ and $F$ are respectively the mid-points of equal sides $AB$ and $AC$ of $\Delta ABC$ (see figure). Show that $BF = CE$.

$l$ and $m$ are two parallel lines intersected by another pair of parallel lines $p$ and $q$ (see Fig). Show that $\Delta ABC \cong \Delta CDA$.

$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.

$AB$ 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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$AD$ is an altitude of an isosceles triangle $ABC$ in which $AB = AC$. Show that:
$(i)$ $AD$ bisects $BC$
$(ii)$ $AD$ 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:
$(i)$ $\Delta DAP \cong \Delta EBP$
$(ii)$ $AD = BE$