Question

Be and CF are two equal altitudes of a triangle ABC using RHS congruence rule prove that triangle​

Answer 1

$\huge\mathbb{\underline{QUESTION:-}}$

Be and CF are two equal altitudes of a triangle ABC using RHS congruence rule prove that triangle .

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$\huge\mathbb{\underline{SOLUTION:-}}$

GIVEN :-

★ BE and CF are two equal altitudes of triangle ABC

★ BE is a altitude

∴ ∠BEC=90 °

And CF is a altitude

∴ ∠BFC=90 °

In ΔBEC and BFC

BE=CF ...Given

BC=BC ...Common

∠BFC=∠BEC=90° ...(Proved above)

∴ ΔBEC ≅ BFC ...RHS test of Congruence

∴ BE=CF

The E and F are mid point of AB and AC

∴ AB = AC

Then, triangle ABC is an isosceles triangle.

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Answer 2

Answer:

Sameer here ✌️✌️✌️

Be and CF are two equal altitudes of a triangle ABC using RHS congruence rule prove that triangle .

____________________________________________________________

GIVEN :-

★ BE and CF are two equal altitudes of triangle ABC

★ BE is a altitude

∴ ∠BEC=90 °

And CF is a altitude

∴ ∠BFC=90 °

In ΔBEC and BFC

BE=CF ...Given

BC=BC ...Common

∠BFC=∠BEC=90° ...(Proved above)

∴ ΔBEC ≅ BFC ...RHS test of Congruence

∴ BE=CF

The E and F are mid point of AB and AC

∴ AB = AC

Then, triangle ABC is an isosceles triangle.

______________________________

♥️♥️ Radhe Radhe ji♥️♥️