Question

In ΔABC, BC is produced to D so that AB = AC = CD. If ∠BAC = 72°, find the angles of ΔABD and arrange its sides in ascending order of length.

Answer 1

Given:

• In ΔABC, BC is produced to D such that AB = AC = CD,
• ∠BAC = 72°

To find:

• Angles of ΔABD,
• Arrange sides of ΔABD in ascending order of length.

Step 1: ΔABC

AB = AC, so ΔABC is isosceles.

∠BAC = 72°.

Therefore,

∠ABC = ∠ACB = `(180^circ - 72^circ)/2 = 54^circ` 

Step 2: ΔACD

Since BC is extended to D, ∠ACD is an exterior angle to ΔABC.

∠ACD = 180° − ∠ACB = 180° − 54° = 126°

In ΔACD, AC = CD (isosceles)

Base angles are equal:

∠CAD = ∠CDA = `(180^circ - 126^circ)/2 = 27^circ`

Step 3: ΔABD

∠ABD = ∠ABC = 54°.

∠CAD = 27°, and ∠BAC = 72°, so

∠BAD = 72° + 27° = 99°

Now,

∠ADB = 180° − (99° + 54°) = 27°

Step 4: Order of sides

• The smallest side is opposite 27° → AB.
• The next side is opposite 54° → AD.
• The largest side is opposite 99° → BD.

So, AB < AD < BD