Iwill award
given: ∆abc, ab = cb
bd - median, e∈ ab
f ∈ bc , ae = cf

△ade≅△cdf by what reason?

1) By SAS theorem, ΔADE≅ΔCDF

2) By SSS theorem, ΔBDE≅ΔBDF

Step-by-step explanation:

Consider isosceles triangle ABC (see diagram).

1. In triangles ADE and CDF:

AD≅DC (since BD is median, then it divides side AC in two congruent parts);

AE≅CF (given);

∠A≅∠C (triangle ABC is isosceles, then angles adjacent to the base are congruent).

By SAS theorem, ΔADE≅ΔCDF.

2. In triangles BDE and BDF:

side BD is common;

DE≅DF (ΔADE≅ΔCDF, then congruent triangles have congruent corresponding sides);

BE≅FB (triangle ABC is isosceles, AB≅BC, AE≅CF, then BE=AB-AE, FB=BC-CF).

Be SSS theorem, ΔBDE≅ΔBDF.

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐