Question

Question 27 pls answer fast

Answer 1

in ∆ADB,
AD = BD
therefore <ABD = <BAD (angles opposite equal sides are equal) ___ (i)
as <ADB + <ABD + <BAD = 180° (angle sum property)
therefore 180° - <ADB = <ABD + <BAD = 2<BAD
so <ADC = 2<BAD

in ∆ACD,
AD = CD
therefore <ACD = <CAD (same as i) ___ (ii)
as <ACD + <CAD + <ADC = 180° (angle sum property)
therefore 180° - <ADC = <ACD + <CAD = 2<CAD
so <ADB = 2<CAD

as <ADB + <ADC = 180°
therefore 2<BAD + 2<CAD = 180°
therefore <BAD + <CAD = 90°
<BAC = 90°