Proof of this theorem?

Question

Let segment AP be an altitude of triangle ABC, and let segment DQ be an altitude of triangle DEF. Prove that if AP=DQ, then the area of ABC / area of DEF = BC / EF. In words, prove that when two triangles have the same altitude, their areas are in the same ratio as their bases.

Answer 1

This isn't nearly as difficult as it seems.

If AP is an altitude, then its corresponding base is BC, so that the area of ABC is

(AP)(BC)/2.

If DQ is an altitude, then its corresponding base is EF, so that the area of DEF is

(DQ)(EF)/2.



Just put the areas on top of each other in a fraction. Change DQ to AP. Then the 2's and AP's cancel, leaving BC/EF.

Answer 2

since the area of a triangle is 1/2 the altitude times the base, triangles with the same altitude have ares in proportion to their bases.

A = 1/2xy
B = 1/2uv
u=x

therefore A/B = y/v

Answer 3

A1 = 1/2x * C
A2 = 1/2y * C

Dividing:
A1/A2 = x/y

Q.E.D.