Question

Pls help!
and a reason.

Answer 1

`\angle BDA = 72^(\circ)`

`\angle DAB = 36^(\circ)`

`\angle BDC = 108^(\circ)`

`\angle BCD = 36^(\circ)`

`\angle DBC = 36^(\circ)`

Explanation:

Given

`\angle ABD = 72^(\circ)`

Required

Determine the missing angles

Since AB = AD, then:

`\angle ABD = \angle BDA` --- Base angles of an isosceles triangle

Hence:

`\angle ABD = \angle BDA = 72^(\circ)`

`\angle ABD + \angle BDA + \angle DAB = 180^(\circ)` --- angles in a triangle

`72^(\circ) + 72^(\circ) + \angle DAB = 180^(\circ)`

`144^(\circ) + \angle DAB = 180^(\circ)`

Collect Like Terms

`\angle DAB = 180^(\circ) - 144^(\circ)`

`\angle DAB = 36^(\circ)`

`\angle BDA + \angle BDC = 180^(\circ)` --- angle on a straight line

`72^(\circ) + \angle BDC = 180^(\circ)`

`\angle BDC = 180^(\circ) - 72^(\circ)`

`\angle BDC = 108^(\circ)`

Since ABC is isosceles, then:

`\angle DAB = \angle BCD = 36^(\circ)` --- base angle of isosceles

`\angle BCD = 36^(\circ)`

Lastly:

`\angle BCD + \angle BDC + \angle DBC = 180^(\circ)`

`36^(\circ) + 108^(\circ) + \angle DBC = 180^(\circ)`

`\angle DBC = 180^(\circ) - 36^(\circ) - 108^(\circ)`

`\angle DBC = 36^(\circ)`

See attachment