Question 4 (ii)
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see the given figure). Show that
AB = AC,
$i . e . Δ A B C i s a n i s o s c e l e s t r i a n g l e .$

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Solution

$I n Δ A E B a n d Δ A F C, W e h a v e ∠ A = ∠ A (C o m m o n a n g l e) B E = C F (G i v e n) ∠ B E A = ∠ C F A (E a c h e q u a l t o 90^{\circ}) T h e r e f o r e, b y A A S r u l e, Δ A E B ≅ Δ A F C$
$W e h a v e p r o v e d a b o v e t h a t Δ A E B ≅ Δ A F C . ∴ A B = A C (C o r r e s p o n d i n g p a r t s o f c o n g r u e n t t r i a n g l e s a r e e q u a l) \Rightarrow Δ A B C i s i s o s c e l e s .$

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Similar questions

i . e . Δ A B C i s a n i s o s c e l e s t r i a n g l e .

$A B C$ is a triangle in which altitudes $B E$ and $C F$ to sides $A C$ and $A B$ are equal (see the given figure). Show that

(i) $△ A B E ≅ △ A C F$

(ii) $A B = A C,$ i.e., $A B C$ is an isosceles triangle.

Δ A B E ≅ Δ A C F

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