Question

write the brief BPT theorem

Answer 1

Two triangles are similar if their corresponding are equal and corresponding sides are proportional.
Basic Proportionality Theorem or Thales Theorem.
If a line is drawn parallel to one side of a triangle, to interest the other two sides indistinct points, the other two sides are divided in the same ratio.

Answer 2

Claim 1 Edit
Intercept theorem proof 2.svg
Since {\displaystyle CA\parallel BD} CA\parallel BD, the altitudes of {\displaystyle \triangle CDA} \triangle CDA and {\displaystyle \triangle CBA} \triangle CBA are of equal length. As those triangles share the same baseline, their areas are identical. So we have {\displaystyle |\triangle CDA|=|\triangle CBA|} |\triangle CDA|=|\triangle CBA| and therefore {\displaystyle |\triangle SCB|=|\triangle SDA|} |\triangle SCB|=|\triangle SDA| as well. This yields

{\displaystyle {\frac {|\triangle SCA|}{|\triangle CDA|}}={\frac {|\triangle SCA|}{|\triangle CBA|}}} {\frac {|\triangle SCA|}{|\triangle CDA|}}={\frac {|\triangle SCA|}{|\triangle CBA|}} and {\displaystyle {\frac {|\triangle SCA|}{|\triangle SDA|}}={\frac {|\triangle SCA|}{|\triangle SCB|}}} {\frac {|\triangle SCA|}{|\triangle SDA|}}={\frac {|\triangle SCA|}{|\triangle SCB|}}

Plugging in the formula for triangle areas ( {\displaystyle {\tfrac {{\text{baseline}}\cdot {\text{altitude}}}{2}}} {\tfrac {{\text{baseline}}\cdot {\text{altitude}}}{2}}) transforms that into

{\displaystyle {\frac {|SC||AF|}{|CD||AF|}}={\frac {|SA||EC|}{|AB||EC|}}} {\frac {|SC||AF|}{|CD||AF|}}={\frac {|SA||EC|}{|AB||EC|}} and {\displaystyle {\frac {|SC||AF|}{|SD||AF|}}={\frac {|SA||EC|}{|SB||EC|}}} {\frac {|SC||AF|}{|SD||AF|}}={\frac {|SA||EC|}{|SB||EC|}}

Canceling the common factors results in:

(a) {\displaystyle \,{\frac {|SC|}{|CD|}}={\frac {|SA|}{|AB|}}} \,{\frac {|SC|}{|CD|}}={\frac {|SA|}{|AB|}} and (b) {\displaystyle \,{\frac {|SC|}{|SD|}}={\frac {|SA|}{|SB|}}} \,{\frac {|SC|}{|SD|}}={\frac {|SA|}{|SB|}}

Now use (b) to replace {\displaystyle |SA|} |SA| and {\displaystyle |SC|} |SC| in (a): {\displaystyle {\frac {\frac {|SA||SD|}{|SB|}}{|CD|}}={\frac {\frac {|SB||SC|}{|SD|}}{|AB|}}} {\frac {\frac {|SA||SD|}{|SB|}}{|CD|}}={\frac {\frac {|SB||SC|}{|SD|}}{|AB|}}

Using (b) again this simplifies to: (c) {\displaystyle \,{\frac {|SD|}{|CD|}}={\frac {|SB|}{|AB|}}} \,{\frac {|SD|}{|CD|}}={\frac {|SB|}{|AB|}} {\displaystyle \,\square } \,\square

Claim 2 Edit
Intercept theorem proof2.svg
Draw an additional parallel to {\displaystyle SD} SD through A. This parallel intersects {\displaystyle BD} BD in G. Then one has {\displaystyle |AC|=|DG|} |AC|=|DG| and due to claim 1 {\displaystyle {\frac {|SA|}{|SB|}}={\frac {|DG|}{|BD|}}} {\frac {|SA|}{|SB|}}={\frac {|DG|}{|BD|}} and therefore {\displaystyle {\frac {|SA|}{|SB|}}={\frac {|AC|}{|BD|}}} {\frac {|SA|}{|SB|}}={\frac {|AC|}{|BD|}}

{\displaystyle \square } \square

Claim 3 Edit
Intercept Theorem - proof 3.svg
Assume {\displaystyle AC} AC and {\displaystyle BD} BD are not parallel. Then the parallel line to {\displaystyle AC} AC through {\displaystyle D} D intersects {\displaystyle SA} SA in {\displaystyle B_{0}\neq B} B_{0}\neq B. Since {\displaystyle |SB|:|SA|=|SD|:|SC|} |SB|:|SA|=|SD|:|SC| is true, we have
{\displaystyle |SB|={\frac {|SD||SA|}{|SC|}}} |SB|={\frac {|SD||SA|}{|SC|}}
and on the other hand from claim 2 we have
{\displaystyle |SB_{0}|={\frac {|SD||SA|}{|SC|}}} |SB_{0}|={\frac {|SD||SA|}{|SC|}}.
So {\displaystyle B} B and {\displaystyle B_{0}} B_{0} are on the same side of {\displaystyle S} S and have the same distance to {\displaystyle S} S, which means {\displaystyle B=B_{0}} B=B_{0}. This is a contradiction, so the assumption could not have been true, which means {\displaystyle AC} AC and {\displaystyle BD} BD are indeed parallel {\displaystyle \square } \square

Claim 4 Edit
Claim 4 can be shown by applying the intercept theorem for two lines.