In $△ A B C$ , $A B = A C$ ; $B E ⊥ A C$ and $C F ⊥ A B$ . Prove that : $B E = C F$ .

Open in App

Solution

In $△ A B C$ , $A B = A C$
So, $∠ B = ∠ C$ $. . . . (i)$ (angles opposite to equal sides are equal)
In $△$ $B E C$ and $B F C,$
$∠ E = ∠ F = 90^{o}$ (given)
$B C = B C$ (common)
$∠ A B C = ∠ A C B$ (From $(i)$ )
Thus, $△ B E C ≅ B F C$ $(A A S$ congruency $)$
Hence, $B E = C F .$

Suggest Corrections

Similar questions

In triangle ABC, AB = AC; BE $⊥$ AC and CF $⊥$ AB. Prove that :

(i) BE =CF

(ii) AF = AE

△ A B C

A B = A C

B E ⊥ A C

C F ⊥ A B

A F = A E

⊥

⊥

△ A B C

A B = A C

B E ⊥ A C

C F ⊥ A B

B E = C F

ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that $Δ A B C$ is isosceles.

Join BYJU'S Learning Program