In the adjoining figure, AB = AC and BD = DC. Prove that ΔADB ≅ ΔADC and hence show that (i) ∠ADB = ∠ADC = 90°, (ii) ∠BAD = ∠CAD. Share with your friends Share 6 Varun Rawat answered this In ∆ADB and ∆ADCAB = AC givenBD = DC given AD = AD common⇒∆ADB ≅ ∆ADC SSS⇒∠ADB = ∠ADC CPCTNow, ∠ADB + ∠ADC = 180° linear pair⇒∠ADB + ∠ADB = 180°⇒∠ADB = 90° = ∠ADCAlso, ∠BAD = ∠CAD CPCT⇒∆ABC is an isosceles ∆. 9 View Full Answer Hema answered this turn the triangle up side then u can see a straight line that is 180. then subtract 180-90=90 therefore u get triangle adb=90 -2