BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule. prove that the triangle ABC is isosceles
Given: BE and CF are two equal altitudes of a triangle ABC
To do: Prove triangle AC is an isosceles using RHS congruence.
Solution:
ABC is a triangle and BE and CF are two altitudes drawn from B and C on to opposite sides AC and AB respectively.
In the right triangles BCF and CBE
BC = BC [hypotenuse]
BE = CF [equal altitudes given]
angle BFC = angle BEC = 90 degrees
So the above triangles are congruent by RHS
So angle B = angle C [by CPCT]
So, the given triangle is an isosceles triangle.
Related Articles
- \( \mathrm{BE} \) and \( \mathrm{CF} \) are two equal altitudes of a triangle \( \mathrm{ABC} \). Using RHS congruence rule, prove that the triangle \( \mathrm{ABC} \) is isosceles.
- AD, BE and CF, the altitudes of ABC are equal. Prove that is an equilateral triangle.
- $ABC$ is a triangle in which $BE$ and $CF$ are respectively, the perpendiculars to the sides $AC$ and $AB$. If $BE = CF$, prove that $\triangle ABC$ is isosceles.
- Angles $A, B, C$ of a triangle $ABC$ are equal to each other. Prove that $\triangle ABC$ is equilateral.
- ABC is an isosceles triangle with AC = BC. If $AB^2 = 2AC^2$. Prove that ABC is a right triangle.
- $ABC$ is a triangle and $D$ is the mid-point of $BC$. The perpendiculars from $D$ to $AB$ and $AC$ are equal. Prove that the triangle is isosceles.
- If $D$ and $E$ are points on sides $AB$ and $AC$ respectively of a $\triangle ABC$ such that $DE \parallel BC$ and $BD = CE$. Prove that $\triangle ABC$ is isosceles.
- ABC is an isosceles triangle right angled at C. Prove that $AB^2 = 2AC^2$.
- $AD$ is an altitude of an equilateral triangle $ABC$. On $AD$ as base, another equilateral triangle $ADE$ is constructed. Prove that Area($\triangle ADE$):Area($\triangle ABC$)$=3:4$.
- In an isosceles $\triangle ABC$, the base AB is produced both the ways to P and Q such that $AP \times BQ = AC^2$. Prove that $\triangle APC \sim \triangle BCQ$.
- \( \mathrm{ABC} \) is a triangle. Locate a point in the interior of \( \triangle \mathrm{ABC} \) which is equidistant from all the vertices of \( \triangle \mathrm{ABC} \).
- $ABC$ is a triangle and through $A, B, C$ lines are drawn parallel to $BC, CA$ and $AB$ respectively intersecting at $P, Q$ and $R$. Prove that the perimeter of $\triangle PQR$ is double the perimeter of $\triangle ABC$.
Kickstart Your Career
Get certified by completing the course
Get Started
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies