Ex.7.3 Q4 Triangles Solution - NCERT Maths Class 9

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

 Video Solution
Ex 7.3 | Question 4

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.


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.


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.

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