Question

F.
E
5. In the given figure, BE and CF are two
equal altitudes of AABC.
Show that (i) AABE = AACF,
(ii) AB = AC.
B​

Answer 1

$\huge \text{Answer}$

BE and CF are two equal altitudes of a triangle ABC . Using RHS congruence rule , prove that the triangle ABC is isosceles .

Given: Given BE is a altitude, So, ∠AEB = ∠CEB= 90∘ Also, CF is a altitude, So, ∠AFC = ∠BFC= ... CBE

∴ ∠FBC= ∠ECB So, ∠ABC = ∠ACB AB = AC So, ∆ABC is an isosceles triangle

Answer 2

$\underline \bold \red \text{Solution}$

BE and CF are two equal altitudes of a triangle ABC . Using RHS congruence rule , prove that the triangle ABC is isosceles .

Given: Given BE is a altitude, So, ∠AEB = ∠CEB= 90∘ Also, CF is a altitude, So, ∠AFC = ∠BFC= ... CBE

∴ ∠FBC= ∠ECB So, ∠ABC = ∠ACB AB = AC So, ∆ABC is an isosceles triangle