Question

In ∆ ABC, AD bisect BC at D. angle ACD = 30° and angle ADC = 45°. Find angle DAB​

Answer 1

Let AH be the altitude of triangle ΔABC through vertex A , i.e. the segment AH is perpendicular to AC with H on AC . The angle ∠BCH=∠BCA=30∘ in the right-angled triangle BCH yields that ∠CBH=60∘ . Furthermore, D is the midpoint of segment BC , so HD is the median of the right angled triangle ΔBCH which allows us to conclude that triangle BDH is equilateral and that BD=CD=HD=HB . Moreover, ∠CHD=30 .

Angle chasing in triangle ΔACD shows that ∠HAD=∠CAD=15∘ . But together with that,

∠HDA=∠HDB−∠ADB=60∘−45∘=15∘

Since ∠HDA=15∘=∠HAD , triangle ΔADH is isosceles with HA=HD . Therefore

HA=HB=HD

Construct the circle centered at point H and radius HA=HB=HD . Then this circle passes through the points A,B and D . From the relation between central and peripheral angles in a circle

∠BAD=12∠BHD=1260∘=30∘