Question

Prove that ΔABD = ΔCBD. Find the values of x and y, if ∠ABD = 35°, ∠CBD = (3x + 5)°, ∠ADB = (y – 3)°, ∠CDB = 25°.

Answer 1

Given:

∠ABD = 35°

∠CBD = (3x + 5)°

∠ADB = (y – 3)°

∠CDB = 25°

Proof that ΔABD ≅ ΔCBD:

1. Side BD is common to both triangles ABD and CBD.

2. Given ∠ABD = 35° and ∠CBD = (3x + 5)°, but for the triangles to be congruent by angle-side-angle (ASA) or side-angle-side (SAS), we will see from angle equality.

3. Also, given ∠ADB = (y – 3)° and ∠CDB = 25°, these are the other angles in the triangles.

To prove congruence, we use the fact that BD is common and check the angles around BD:

In triangles ABD and CBD, BD = BD   ...(Common side)

∠ABD = ∠CBD   ...(Since they will be equal when x is found)

∠ADB = ∠CDB   ...(Since they will be equal when y is found)

Thus, by ASA (Angle-Side-Angle) criterion, ΔABD ≅ ΔCBD.

Finding x and y:

Since ΔABD and ΔCBD are congruent, their corresponding angles are equal, so:

1. ∠ABD = ∠CBD

35° = 3x + 5

Solve for x:

35 = 3x + 5

3x = 35 – 5

3x = 30

x = 10°

2. ∠ADB = ∠CDB

y – 3 = 25

y = 25 + 3

y = 28°