△ABCand△DBCare two isosceles triangles on the same baseBCand verticesAandDare on the same side ofBC. IfADis extended to intersectBCatP, show that(i)△ABD≅△ACD(ii)△ABP≅△ACP(iii)APbisects∠Aas well as△D.(iv)APis the perpendicular bisector ofBC.△ABCand△DBCare two isosceles triangles on the same baseBCand verticesAandDare on the same side ofBC. IfADis extended to intersectBCatP, show that(i)△ABD≅△ACD(ii)△ABP≅△ACP(iii)APbisects∠Aas well as△D.(iv)APis the perpendicular bisector ofBC.△ABC△ABC△ABC△△AABBCC△DBC△DBC△DBC△△DDBBCCBCBCBCBBCCAAAAADDDDDBCBCBCBBCCADADADAADDBCBCBCBBCCPPPPP△ABD≅△ACD△ABD≅△ACD△ABD≅△ACD△△AABBDD≅≅≅△△AACCDD△ABP≅△ACP△ABP≅△ACP△ABP≅△ACP△△AABBPP≅≅≅△△AACCPPAPAPAPAAPP∠A∠A∠A∠∠AA△D△D△D△△DDAPAPAPAAPPBCBCBCBBCC?

(i) In△ABDand△ACD,AB=AC(since△ABCis isosceles)AD=AD(common side)BD=DC(since△BDCis isosceles)ΔABD≅ΔACD.SSS test of congruence,∴∠BAD=∠CADi.e.∠BAP=∠PAC.[c.a.c.t]. (i)(ii) In△ABPand△ACP,AB=AC.. (since△ABCis isosceles)AP=AP.. (common side)∠BAP=∠PACfrom (i)△ABP≅△ACP SAS test of congruence∴BP=PC...[c.s.c.t] (ii)∠APB=∠APCc.a.c.t (iii) Since△ABD≅△ACD∠BAD=∠CADfrom (i)So,ADbisects∠Ai.e.APbisects∠A (iii)In△BDPand△CDP,DP=DP...common sideBP=PC...from (ii)BD=CD.. (since△BDCis isosceles)△BDP≅△CDPSSS test of congruence∴∠BDP=∠CDPc.a.c.t.∴DPbisects∠DSo,APbisects∠D(iv)From (iii) and (iv),APbisects∠Aas well as∠D (iv) We know that∠APB+∠APC=180∘(angles in linear pair)Also,∠APB=∠APC...from (ii)∴∠APB=∠APC=180∘2=90∘BP=PCand∠APB=∠APC=90∘Hence,APis perpendicular bisector ofBC.