Question

AD,BE and CF are altitudes of triangle are equal. Find 2 right angles with RHS criteria.

Please answer with explanation.
I'll mark brainliest.

use RHS CRITERIA​

Answer 1

Step-by-step explanation:

In right-angle triangles BCE and CBF, we have,

BC = BC (common hypotenuse);

BE = CF (given).

Hence BCF and CBF are congruent, by RHS theorem. Comparing the triangles, we get ∠B=∠C.

This implies that

AC = AB (sides opposite to equal angles).

Similarly,

AD=BE⇒∠B=∠A

⇒AC=BC

Together, we get AB=BC=ACor △ABC is equilateral. [henceproved]

Answer 2

Step-by-step explanation:

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

$hope \: \: \: \: it \: \: \: \: \: will \: \: help \: \: \: you$