Question

Please give correct answer...​

Answer 1

This is the solution of the problem.

Answer 2

Answer:

Given :

(a)

• $\ AD \ bisects \ \angle A$
• $DE \perp CA$
• $DF \perp AB$

(b)

• $\angle ABE = \angle ACF$
• $AB = AC$

Solution:

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(a)

$In \triangle ADF \ and \ \triangle ADE$

1️⃣ $\angle AFC = \angle AED \ \ (each \ 90^0)$

2️⃣ $\angle FAD = \angle EAD \ \ (AD \ is \ bisector)$

3️⃣ $AD = AD \ \ (common)$

$\sf{\boxed{By \ AAS \ property \ \triangle ADF \cong \triangle ADE}}$

$\implies AF = AE \ (C.P.C.T.)$

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(b)

$In \triangle ABE \ and \ \triangle ACF$

1️⃣ $\angle ABE = \angle ACF \ \ (given)$

2️⃣ $\angle A = \angle A \ \ (common)$

3️⃣ $AB = AC \ \ (given)$

$\sf{\boxed{By \ AAS \ property \triangle ABE \cong \triangle ACF}}$

$\implies BE = CF \ (C.P.C.T.)$

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