Question 46

Given:

∠A₁ = ∠A₂ AB ⊥ BC AD ⊥ DC

(AC bisects angle A, and the angles at B and D are right angles)

a. Prove: ΔABC ≅ ΔADC

Step 1: Identifying the right angles

From the given AB ⊥ BC and AD ⊥ DC, what are the angle values?

∠ABC = ° ∠ADC = °

Correct! Both angles are 90°

Step 2: Identifying the common hypotenuse

Both triangles are right triangles. What is the common hypotenuse?

Common hypotenuse:

Correct! AC is the common hypotenuse

Step 3: Identifying the equal acute angle

From the given ∠A₁ = ∠A₂, what are the equal angles in the triangles?

∠BAC = ∠

Correct! ∠BAC = ∠DAC (or ∠CAD)

Step 4: Choosing the congruence theorem

We have right triangles with a common hypotenuse and an equal acute angle.

What is the appropriate congruence theorem?

Excellent! Hypotenuse-Acute Angle is the correct theorem!

Proof summary:
• ∠ABC = 90° and ∠ADC = 90° (right angles from the given)
• AC is the common hypotenuse in both right triangles
• ∠BAC = ∠DAC (since ∠A₁ = ∠A₂, AC bisects angle A)
Therefore by Hypotenuse-Acute Angle → ΔABC ≅ ΔADC

b. Prove: AB = AD

From the congruence we proved, is it true that AB = AD?

In congruent triangles, corresponding sides are equal.
From the congruence ΔABC ≅ ΔADC, does AB correspond to AD?

Correct! AB = AD because they are corresponding sides in congruent triangles!

From the congruence ΔABC ≅ ΔADC:
The correspondence is: A↔A, B↔D, C↔C
Therefore the corresponding sides are:
• AB = AD (first sides)
• BC = DC (second sides)
• AC = AC (third sides - shared)

Conclusion: AB = AD

Well done! You finished the question!