$BE$ and $CF$ are two equal altitudes of a triangle $ABC$. Using $RHS$ congruence rule,prove that the triangle $ABC$ is isosceles.
(N/A) Given: $BE \perp AC$ and $CF \perp AB$ such that $BE = CF$.
To prove: $\Delta ABC$ is an isosceles triangle,i.e.,$AB = AC$.
Proof:
In right-angled triangles $\Delta BEC$ and $\Delta CFB$:
$1$. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes are perpendicular to the sides).
$2$. $BC = CB$ (Common hypotenuse).
$3$. $BE = CF$ (Given).
Therefore,by $RHS$ congruence rule,$\Delta BEC \cong \Delta CFB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$\angle BCE = \angle CBF$
This implies $\angle BCA = \angle CBA$.
In $\Delta ABC$,since the angles opposite to equal sides are equal,the sides opposite to equal angles are also equal.
Therefore,$AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.
To prove: $\Delta ABC$ is an isosceles triangle,i.e.,$AB = AC$.
Proof:
In right-angled triangles $\Delta BEC$ and $\Delta CFB$:
$1$. $\angle BEC = \angle CFB = 90^\circ$ (Altitudes are perpendicular to the sides).
$2$. $BC = CB$ (Common hypotenuse).
$3$. $BE = CF$ (Given).
Therefore,by $RHS$ congruence rule,$\Delta BEC \cong \Delta CFB$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$:
$\angle BCE = \angle CBF$
This implies $\angle BCA = \angle CBA$.
In $\Delta ABC$,since the angles opposite to equal sides are equal,the sides opposite to equal angles are also equal.
Therefore,$AB = AC$.
Hence,$\Delta ABC$ is an isosceles triangle.
Explore More
Similar Questions
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 $\angle A$.
$(i)$ $\Delta APB \cong \Delta AQB$
$(ii)$ $BP = BQ$ or $B$ is equidistant from the arms of $\angle A$.
Medium
View Solution$AD$ and $BC$ are equal perpendiculars to a line segment $AB$ (see figure). Show that $CD$ bisects $AB$.
Medium
View Solution$ABC$ is an isosceles triangle with $AB = AC$. Draw $AP \perp BC$ to show that $\angle B = \angle C$.
Medium
View Solution$P$ is a point equidistant from two lines $l$ and $m$ intersecting at point $A$. Show that the line $AP$ bisects the angle between them.
Medium
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 SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo