The altitudes of equal length, BE and CF gives AB = AC, from which we have;
- ∆ABC is an isosceles triangle
How can the RHS rule be used to indicate an isosceles triangle?
From the given description, we have;
BE = CF
<CFB = <BEC = 90° All right angles are congruent
BC = BC by reflexive property of congruency
BC = The hypotenuse side of triangles ∆BFC and ∆CEB
Therefore;
∆BFC is congruent to ∆CEB by Right angle Hypotenuse Side, RHS, rule of congruency.
Therefore;
BF = CE by Corresponding Parts of Congruent Triangles are Congruent, CPCTC
Similarly, we have;
∆AEB is congruent to ∆AFC, by Side-Angle-Angle, SAA, rule of congruency
Which gives;
FA = AE by CPCTC
BF + FA = CE + AE, by substitution property of equality
BF + FA = AB
CE + AE = AC
Therefore;
Therefore;
- ∆ABC is an isosceles triangle
Learn more about rules of congruency here:
https://brainly.com/question/24261247
#SPJ1
