Given: ∆abc, ab = cb bd − median to ac e∈ ab ,f∈ bc ae = cf prove: △ade ≅ △cdf δbde ≅ δbdf

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:

By SAS theorem, ΔADE≅ΔCDF.

2. In triangles BDE and BDF:

Be SSS theorem, ΔBDE≅ΔBDF.

c. 27 units.

step-by-step explanation:

we are told that in circle a, ∠bae ≅ ∠dae.

we can see from our given diagram that in and ;

, as these are radii of our given circle.

therefore, by sas congruence postulate.

hence, side be will be equal to side de as corresponding parts of congruent triangles are congruent. so we can set an equation to find the value of x as:

now let us substitute x=17 in the expression for the length of be.

therefore, length of be will be 27 units and option c is the correct choice.