$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal (see Fig). Show that
$(i)$ $\Delta ABE \cong \Delta ACF$
$(ii)$ $AB = AC$,i.e.,$ABC$ is an isosceles triangle.
$(i)$ $\Delta ABE \cong \Delta ACF$
$(ii)$ $AB = AC$,i.e.,$ABC$ is an isosceles triangle.
(N/A) $(i)$ In $\Delta ABE$ and $\Delta ACF$,we have:
$\angle AEB = \angle AFC = 90^\circ$ (Since $BE \perp AC$ and $CF \perp AB$)
$\angle A = \angle A$ (Common angle)
$BE = CF$ (Given)
Therefore,by the $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
$(ii)$ Since $\Delta ABE \cong \Delta ACF$,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Thus,$ABC$ is an isosceles triangle.
$\angle AEB = \angle AFC = 90^\circ$ (Since $BE \perp AC$ and $CF \perp AB$)
$\angle A = \angle A$ (Common angle)
$BE = CF$ (Given)
Therefore,by the $AAS$ congruence criterion,$\Delta ABE \cong \Delta ACF$.
$(ii)$ Since $\Delta ABE \cong \Delta ACF$,their corresponding parts are equal $(CPCT)$.
Therefore,$AB = AC$.
Thus,$ABC$ is an isosceles triangle.
Explore More
Similar Questions
Show that of all line segments drawn from a given point not on a line,the perpendicular line segment is the shortest.
Difficult
View Solution$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$ (see figure). Show that $\angle ABD = \angle ACD$.
Medium
View Solution$ABC$ is a triangle. Locate a point in the interior of $\Delta ABC$ which is equidistant from all the vertices of $\Delta ABC$.
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 Solution$ABCD$ is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see figure). Prove that:
$(i)$ $\Delta ABD \cong \Delta BAC$
$(ii)$ $BD = AC$
$(iii)$ $\angle ABD = \angle BAC$
$(i)$ $\Delta ABD \cong \Delta BAC$
$(ii)$ $BD = AC$
$(iii)$ $\angle ABD = \angle BAC$
Medium
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