State BPT theorem and prove it.

Hint: To prove this theorem first we will join BE and CD. Then draw a line EL perpendicular to AB and line DM perpendicular to AC. Now we will find the ratio of area of Δ ADE to Δ DBE and ratio of area of Δ ADE to Δ ECD. Comparing the ratios we will get the final answer. Complete step-by-step answer: Now, ΔDBE and ΔECD being on the same base DE and between the same parallels DE and BC, we have,ar(ΔDBE)=ar(ΔECD) then we say that the basic proportionality theorem is proved. Basic proportionality theorem: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points then the other two sides are divided in the same ratio. Given: A ΔABC in which DE∥BC and DE intersects AB and AC at D and E respectively. To prove that: ADDB=AEEC Construction: Join BE and CD. Draw EL⊥AB and DM⊥AC Proof: We have the ar(ΔADE)=12×AD×EL ar(ΔDBE)=12×DB×EL Therefore the ratio of these two is ar(ΔADE)ar(ΔDBE)=ADDB . . . . . . . . . . . . . . (1) Similarly, ar(ΔADE)=ar(ΔADE)=12×AE×DM ar(ΔECD)=12×EC×DM Therefore the ratio of these two is ar(ΔADE)ar(ΔECD)=AEEC . . . . . . . . . . . .. . . (2) Now, ΔDBE and ΔECD being on the same base DE and between the same parallels DE and BC, we have, ar(ΔDBE)=ar(ΔECD) . . . . . . . . . . . (3) From equations 1, 2, 3 we can conclude that ADDB=AEEC Hence we can say that the basic proportionality theorem is proved. Note: The formula for area of the triangle is given by 12×b×h where b, h are base and height respectively. If two triangles are on the same base and between the same parallels then the area of those two triangles are equal.