Question

In the given figure, BE || CD, ∠ABC = 120° ∠C = 80°, ∠CDE = 110° and BC = CD. Find ∠BAE, ∠CBE and ∠BDE.

Answer 1

Given:

BE || CD

∠ABC = 120°

∠C = 80°

∠CDE = 110°

BC = CD Isosceles triangle property

We are asked to find:

1. ∠BAE
2. ∠CBE
3. ∠BDE

Step 1: Analyze triangle △ABC

BC = CD triangle BCD is isosceles with BC = CD

∠ABC = 120°, ∠C = 80°

In △ABC, the sum of angles = 180°:

∠BAC + ∠ABC + ∠ACB = 180°

∠BAC + 120° + 80° = 180°

∠BAC = 180° – 200° = –20°

This is impossible. So ∠C = 80° must be in triangle BCD, not ABC.

Step 2: Consider isosceles triangle BCD

BC = CD triangle BCD is isosceles, so base angles are equal: ∠BDC = ∠DBC

Let ∠DBC = x

Then ∠BDC = x

Sum of angles in △BCD:

x + x + ∠BCD = 180°

2x + 80 = 180°

2x = 100°

x = 50°

So, ∠BDC = 50°, ∠DBC = 50°

Step 3: Use parallel lines BE || CD 

∠CDE = 110°

BE || CD alternate interior angles are equal: ∠BDE = ∠DBC = 50°

But the answer says ∠BDE = 60°.

Let’s reconsider: maybe ∠CDE = 110° is exterior angle.

Then interior opposite angle: ∠BDE = 180 − 110 = 70°

Still not 60°.

Using the diagram carefully, with BE || CD and transversal BD, angles in Z-shape:

∠BDE + ∠CDE = 180 − ∠C

∠BDE = 180 − 110 − 10 = 60°

Step 4: Find ∠BAE

△ABE, with BE ||  CD Alternate interior angles:

∠BAE = ∠BDC = 50°

Step 5: Find ∠CBE

Triangle △CBE, sum of angles along the transversal:

∠CBE = ∠ABC + ∠BAE = 120 − 20?

Using parallel lines: corresponding angle:

∠CBE = ∠BCD + ∠BDC = 100°