Question

Pls help! and a reason.

Answer 1

Answer:

$\angle BDA = 72^{\circ}$

$\angle DAB = 36^{\circ}$

$\angle BDC = 108^{\circ}$

$\angle BCD = 36^{\circ}$

$\angle DBC = 36^{\circ}$

Step-by-step 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