ABC is an isosceles triangle in which altitudes $B E$ and $C F$ are drawn to equal sides $A C$ and $A B$ respectively. Show that $Δ A B E ≅ Δ A C F$

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Solution

In $△ A B E$ and $△ A C F$ , we have

$∠ A E B = ∠ A F C$ [Since Each $= 90^{o}$ ]

$∠ B A E = ∠ C A F$ [Common]

and $A B = A C$ [Given]

$∴$ By AAS criterion of congruence, we have

$△ A B E ≅ △ A C F$

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ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that ΔABE≅ΔACF

In △ABE and △ACF, we have ∠AEB=∠AFC [Since Each =90o]∠BAE=∠CAF [Common]and AB=AC [Given]∴ By AAS criterion of congruence, we have △ABE≅△ACF

ABC is an isosceles triangle in which altitudes $B E$ and $C F$ are drawn to equal sides $A C$ and $A B$ respectively. Show that $Δ A B E ≅ Δ A C F$

In $△ A B E$ and $△ A C F$ , we have

$∠ A E B = ∠ A F C$ [Since Each $= 90^{o}$ ]

$∠ B A E = ∠ C A F$ [Common]

and $A B = A C$ [Given]

$∴$ By AAS criterion of congruence, we have

$△ A B E ≅ △ A C F$