Question

BE and CF are two equal altitudes of a triangle ABC. using RHS congruence rule prove that ABC is isosceles​

Answer 1

In ΔBCF andΔCBE,

∠BFC=∠CEB (Each 90º)

Hyp. BC= Hyp. BC (Common Side)

Side FC= Side EB (Given)

∴ By R.H.S. criterion of congruence, we have

ΔBCF≅ΔCBE

∴∠FBC=∠ECB (CPCT)

In ΔABC,

∠ABC=∠ACB

[∵∠FBC=∠ECB]

∴AB=AC (Converse of isosceles triangle theorem)

∴ ΔABC is an isosceles triangle.

Answer 2

Given -

• BE and CF are 2 equal altitudes.

To find -

• That ∆ABC is an isosceles triangle.

Solution -

In ∆BCF and ∆CBE

→ $\angle$ BFC = $\angle$ CBE (Both 90°)

→ BC = BC (common)

→ FC = CB (Given)

$\therefore$ ∆BCF ≈ ∆CBE (RHS rule)

So,

$\angle$FCB = $\angle$ ECB (CPCT)

Similarly -

AB = AC (Side opposite to the equal angles of a triangle are also equal)

$\therefore$ ∆ABC is an isosceles triangle.

Hence, proved!

Some more rules of a Triangle -

• ASA rule

When 2 angles and one side of a Triangle are equal to the 2 angles and one side of another Triangle. Then Triangles are congurent.

• SAS rule

When 2 sides and one angle of a triangle are equal to the 2 sides and one angle of another Triangle. Then Triangles are congurent.

• SSS rule

If 3 sides of a Triangle are equal to the 3 sides of another Triangle. Then Triangles are congurent.

• RHS rule

If in 2 right triangles the hypotenuse and one side of one Triangle are equal to the hypotenuse and one angle of another Triangle. Then Triangles are congurent.

• AAS rule

When 2 angles and one sides are equal to 2 angles and one side of a Triangle. Then Triangles are congurent.

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