Activity:In AABC, ray BD bisects ZABC.A-D-C, side DE || side BC, A-E-B

1.	Activity:In AABC, ray BD bisects ZABC.A-D-C, side DE \|\| side BC, A-E-B thenprove that, AB AEВСЕВFig. 1.32
Answer» FL hjvfdx Subho still fun my xi mg DG my GVC fh ki gghgg GB HDD gh CMP DMP dp is b skywalk you have a great day and I GIVEN: Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) Plz like my answer and accepted as best plz CMP DMP dp is b sky walk you have agreat day andI If ∆ABD~∆ACE ,then we can prove AB/AC=AD/AE..but AB=AC .....1) AD=AE...2) AB-AE=AC-AD, that employes BE=DC...3)by 2&3 ED//BC... GIVEN: Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ……(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) Plz like my answer and accepted as best plz GIVEN: Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) : Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) Plz like my answer and accepted as best plz AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC jdjdjnnxndnbdbhhnnfbhrjj i dont know answer please send me correct option

Activity:In AABC, ray BD bisects ZABC.A-D-C, side DE || side BC, A-E-B thenprove that, AB AEВСЕВFig. 1.32

Answer»

FL hjvfdx Subho still fun my xi mg DG my GVC fh ki gghgg GB HDD gh

CMP DMP dp is b skywalk you have a great day and I

GIVEN: Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) Plz like my answer and accepted as best plz

CMP DMP dp is b sky walk you have agreat day andI

If ∆ABD~∆ACE ,then we can prove AB/AC=AD/AE..but AB=AC .....1) AD=AE...2) AB-AE=AC-AD, that employes BE=DC...3)by 2&3 ED//BC...

GIVEN: Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ……(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) Plz like my answer and accepted as best plz

: Triangle ABC , AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC ( by converse of Basic proportionality theorem) Plz like my answer and accepted as best plz

AB = AC => < B =< ABD = < ACD In triangle ABD & triangle ACE < ABD = < ACE ( proved above)tri ABD ~ tri ACE ( by AA similarity criterion) => AB/AC = AD/AE ( csst) But AB = AC ( given) …….. (1) => AD = AE ………(2) => AB - AE = AC - AD => BE = DC …….. (3) By (2) & (3) ED // BC

Activity:In AABC, ray BD bisects ZABC.A-D-C, side DE || side BC, A-E-B thenprove that, AB AEВСЕВFig. 1.32

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