Please answer question no.10. Share with your friends Share 0 Priyanka Sahu answered this Dear Student 10.In ∆ABD and ∆ACD,AB = AC givenBD = CD as, D is the mid point of BCAD = AD common⇒∆ABD ≅ ∆ACD SSSNow, ∠BAD = ∠CAD CPCTSo, AD is the bisector of ∠A.Also, ∠ADB = ∠ADC CPCTNow, ∠ADB + ∠ADC = 180° Linear pair⇒∠ADB + ∠ADB = 180° as, ∠ADB = ∠ADC⇒∠ADB = 90° = ∠ADCSo, AD⊥BC Regards 0 View Full Answer
