Proved Thales theorem.

Let ABC be the triangle. The line l parallel to BC intersect AB at D and AC at E. To prove AD​/DB =AE​/EC Join BE,CD Draw EF⊥AB, DG⊥CA Since EF⊥AB, EF is the height of triangles ADE and DBE Area of △ADE=1/2 × base × height=1/2​AD×EF Area of △DBE= 1/2 ​×DB×EF areaofΔDBEareaofΔADE​=1/2​×AD×EF/ 1/2​×DB×EF​=AD/DB​ ........(1) Similarly, areaofΔDCEareaofΔADE​= 1/2 ×AE×DG/ 1/2 ×EC×DG ​=AE/EC​ ......(2) But ΔDBE and ΔDCE are the same base DE and between the same parallel straight line BC and DE. Area of ΔDBE= area of ΔDCE ....(3) From (1), (2) and (3), we have AD​/DB =AE​/EC Hence proved.

Basic Proportionality Theorem states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points,then the line divides those sides of the triangle in proportion.

Let ABC be the triangle.
The line l parallel to BC intersect AB at D and AC at E.
To prove AD​/DB =AE​/EC
Join BE,CD

Draw EF⊥AB, DG⊥CA
Since EF⊥AB,
EF is the height of triangles ADE and DBE
Area of △ADE=1/2 × base × height=1/2​AD×EF
Area of △DBE= 1/2 ​×DB×EF
areaofΔDBEareaofΔADE​=1/2​×AD×EF/ 1/2​×DB×EF​=AD/DB​          ........(1)
Similarly,
areaofΔDCEareaofΔADE​= 1/2 ×AE×DG/ 1/2 ×EC×DG ​=AE/EC​            ......(2)

But ΔDBE and ΔDCE are the same base DE and between the same parallel straight line BC and DE.
Area of ΔDBE= area of ΔDCE         ....(3)
From (1), (2) and (3), we have

AD​/DB =AE​/EC

Hence proved.