Question

(a) The diagonal BD of a parallelogram ABCD intersects the line segment AE at the point F, where E is any point on the side BC. Prove that $D F \times EF = FB \times F A$ .

(b) In $△ A BC$ , if $A D ⊥ BC$ and $A D^{2} = B D \times D C$ , then prove that $∠ B A C = 90°$ .

Approach

In parallelogram $A BC D$ , diagonal $B D$ intersects line segment $A E$ at $F$ , where $E$ is on side $BC$ . By alternate interior angles, $∠ D F A = ∠ EFB$ (vertical) and $∠ A D F = ∠ FBE$ (alternate). In triangles $A D F$ and $EBF$ , we have two pairs of equal angles, so $△ A D F \sim △ EBF$ . By corresponding sides of similar triangles, $\frac{D F}{FB} = \frac{F A}{EF}$ , hence $D F \times EF = FB \times F A$ .

Step-by-step working

1
Draw correct figure showing parallelogram $A BC D$ , diagonal $B D$ , point $F$ on $A E$ where $E$ is on $BC$
1 mark
2
In $△ A D F$ and $△ EBF$ : $∠ D F A = ∠ EFB$ (vertically opposite)
1 mark
3
$∠ A D F = ∠ FBE$ (alternate interior angles, $A D ∥ BC$ )
1 mark
4
$∴ △ A D F \sim △ EBF$ (AA similarity)
1 mark
5
$\frac{D F}{FB} = \frac{F A}{EF} \Rightarrow D F \times EF = FB \times F A$
1 mark

Concepts used

similar trianglesproveparallelogramaltitudepythagorean theoremright angle

Approach

Step-by-step working

Concepts used

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