Question

Prove the triangle proportionality Theorem.

Answer 1

Answer:

Statement of triangle proportionality:  

If a line parallel to one side of a triangle intersects the other two sides of the triangle, then that line divides these two sides proportionally.

From the statement: If $FG || BC$ then,

Show that: $\frac{FB}{FA} = \frac{GC}{AG}$

Consider $\triangle ABC$ and $\triangle GFA$

Reflexive property states that the value is equal to itself.

$\angle BAC \cong \angle GAF$ [Angle] {Reflexive property of equality}

Corresponding angles theorem states that if the two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent(i., e equal).

$\angle ABC \cong \angle GFA$ [Angle]

$\angle ACB \cong \angle AGF$ [Angle]

AA Similarity states that the two triangles have their corresponding angles equal if and only if their corresponding sides are proportional.

then, by AA similarity theorem:  

$\triangle ABC \sim \triangle GFA$

By segment addition postulates:

AB = FA +FB and AC = AG + GC  

Corresponding sides in similar triangles are proportional  

$\frac{AB}{FA} = \frac{AC}{AG}$ .....[1]

Substitute AB = FA +FB and AC = AG + GC in [1]  

we have;

$\frac{FA+FB}{FA} = \frac{AG+GC}{AG}$  

Separate the fraction:

$\frac{FA}{FA} + \frac{FB}{FA} = \frac{AG}{AG} + \frac{GC}{AG}$

Simplify:

$1 + \frac{FB}{FA} =1+ \frac{GC}{AG}$

Subtract 1 from both sides we get;

$\frac{FB}{FA} =\frac{GC}{AG}$ hence proved