Question

in the given figure, if BE = CF, then proof that :-
(a) ∆ABE congruent to ∆ACF (b) ∆ABE congruent to ∆AFC
(c) ∆ABE congruent to ∆CAF​

Answer 1

Answer:

a) since BE perpendicular to AC

therefore angle AEB is also 90°

and CF perpendicular to AB

therefore angle AFC is also 90°

in triangle ABE and ACF

AEB= AFC( each 90°)

angle A = angle A ( common)

BE= CF ( Given)

therefore triangle ABE is congruent to triangle ACF

b) it is part (a) only but alphabetical order have been changed

c) BRO! these all triangles are same only there is change alphabetical arrangements. You don't believe me ask to your teacher.

Hope this will help;)

pls follow I take to much to time to type it pls;)

Answer 2

Concept introduction:

A polygon having three edges and three vertices is called a triangle. It is one of the fundamental geometric forms. Triangle $ABC$ is the designation for a triangle with vertices $A$, $B$, and $C$. In Euclidean geometry, any three points that are not collinear produce a distinct triangle and a distinct plane.

Given:

Here in the following figure it is given that $BE = CF$.

To find:

We have to find that  ∆$ABE$ congruent to ∆$ACF$.

Solution:

According to the question,  $BE = CF$.

since $BE$ perpendicular to $AC$

Therefore angle ∠$AEB$ is also $90$° and $CF$ perpendicular to $AB$   Therefore angle ∠$AFC$ is also $90$° in triangle $ABE$ and $ACF$

$AEB= AFC$( each $90$°)

angle ∠$A$ $=$ angle ∠$A$ ( common)

$BE= CF$( Given)

Therefore triangle $ABE$ is congruent to triangle $ACF$.

Final answer:

So, we have given the proof of triangle $ABE$ is congruent to triangle $ACF$ and this is our final answer also.

#SPJ3