# Ex.7.3 Q4 Triangles Solution - NCERT Maths Class 9

## Question

\(BE\) and \(CF\) are two equal altitudes of a triangle \(ABC\). Using RHS congruence rule, prove that the triangle \(ABC\) is isosceles.

## Text Solution

**What is Known?**

\(BE\) and \(CF\) are two equal altitudes of a triangle \(ABC.\)

**To prove:**

Triangle \(ABC\) is isosceles by using RHS congruence rule.

**Reasoning:**

We can show triangles \(BEC\) and \(CFB\) congruent by using RHS congruency and then we can say corresponding parts of congruent triangles are equal to prove the required result.

**Steps:**

In \(\Delta BEC\) and \(\Delta CFB\),

\[\begin{align} \angle BEC & = \angle CFB\;(\text{Each }90^{\circ})\\ BC &= CB\;(\text{Common})\\ BE &= CF\;(\text{Given})\\ \therefore \Delta BEC &\cong \Delta CFB\\ &\text{(By RHS congruency)} \end{align}\]

\(\therefore \angle BCE = \angle CBF (\text{By }CPCT)\)

\(\therefore AB = AC\) (Sides opposite to equal angles of a triangle are equal) Hence, \(\Delta ABC\) is isosceles.