Question

a student is to show that triangle adb is congruent to triangle adc. ad ⊥ bc, ab ≅ ac. statements: 1. ad ⊥ bc (given) 2. m∠adb = 90°, m∠adc = 90° (definition of perpendicular lines) 3. ∠adb ≅ ∠adc (all right angles are congruent) 4. ab ≅ ac (given) 5. △abc is isosceles (definition of isosceles triangle) 6. ∠abd ≅ ∠acd (base angles of isosceles triangle are congruent) 7. △adb ≅ △adc (aas congruence theorem)

Explanation:

Step1: Analyze Given Information

We know \( AD \perp BC \), so \( \angle ADB = \angle ADC = 90^\circ \) (perpendicular lines form right angles). Also, \( AD \) is common to both \( \triangle ADB \) and \( \triangle ADC \), and from the diagram (or given \( AB = AC \) as a statement? Wait, statement 4 is \( AB \cong AC \)). Wait, let's list the parts:

• \( \angle ADB = \angle ADC = 90^\circ \) (from \( AD \perp BC \), statement 2's reason? Wait, statement 2: \( m\angle ADB = 90^\circ, m\angle ADC = 90^\circ \), reason is "perpendicular lines form right angles".
• \( AD = AD \) (common side, reflexive property).
• \( AB = AC \) (statement 4, given? Or from the problem's "Given" section: \( AB > AC \)? Wait, no, the "Given" is \( AD \perp BC \), \( AB \cong AC \)? Wait, the left side says "Given: \( AD \perp BC \), \( AB \cong AC \)"? Wait, the user's image: "Given: \( AD \perp BC \), \( AB \cong AC \)" and we need to prove \( \triangle ADB \cong \triangle ADC \).

Step2: Identify Congruence Criterion

For right triangles, if hypotenuse and one leg are equal, HL (Hypotenuse-Leg) congruence. But here, \( AD \) is a leg, \( AB = AC \) (hypotenuse), and \( \angle ADB = \angle ADC = 90^\circ \). So \( \triangle ADB \cong \triangle ADC \) by HL (Hypotenuse-Leg) or SAS (since \( AD = AD \), \( \angle ADB = \angle ADC \), \( AB = AC \)? Wait, no, SAS would be two sides and included angle. Here, \( AD \) is a side, \( \angle ADB \) is the angle, and \( AB \) is the hypotenuse. Wait, actually, in right triangles, HL: hypotenuse (\( AB = AC \)) and leg (\( AD = AD \)) so \( \triangle ADB \cong \triangle ADC \) by HL.

Wait, the statements:

1. \( AD \perp BC \) (Given)
2. \( m\angle ADB = 90^\circ, m\angle ADC = 90^\circ \) (Definition of perpendicular lines)
3. \( AD = AD \) (Reflexive Property)
4. \( AB \cong AC \) (Given? Or from the problem's "Given" section)
5. So \( \triangle ADB \cong \triangle ADC \) by HL (Hypotenuse-Leg) congruence, since both are right triangles, hypotenuse \( AB = AC \), leg \( AD = AD \).

Answer:

To prove \( \triangle ADB \cong \triangle ADC \):

1. Given: \( AD \perp BC \), so \( \angle ADB = \angle ADC = 90^\circ \) (right angles).
2. Reflexive Property: \( AD = AD \) (common side).
3. Given (or from diagram): \( AB \cong AC \) (hypotenuse of right triangles).
4. By the Hypotenuse - Leg (HL) Congruence Criterion for right triangles, \( \triangle ADB \cong \triangle ADC \).

(If using SAS: \( AD = AD \), \( \angle ADB = \angle ADC \), \( AB = AC \) – but wait, \( AB \) and \( AC \) are not the sides adjacent to \( \angle ADB \) and \( \angle ADC \). So HL is correct here.)

So the reason for \( \triangle ADB \cong \triangle ADC \) is Hypotenuse - Leg (HL) Congruence, using \( AD \perp BC \) (right angles), \( AD = AD \) (common leg), and \( AB = AC \) (hypotenuse).