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 $A E$ (where $E$ is on $BC$ ) at $F$ . Prove $D F \times EF = FB \times F A$ . Show $△ A D F \sim △ EBF$ using alternate angles and vertically opposite angles, then corresponding sides are proportional.

Step-by-step working

1
Draw correct figure with parallelogram $A BC D$ , point $E$ on $BC$ , diagonals intersecting at $F$ .
1 mark
2
In $△ A D F$ and $△ EBF$ : $∠ D F A = ∠ EFB$ (vertically opposite), $∠ A D F = ∠ FBE$ (alternate angles, $A D ∥ BC$ )
2 marks
3
$△ A D F \sim △ EBF$ (AA similarity)
1 mark
4
$\frac{D F}{FB} = \frac{F A}{EF} \Rightarrow D F \times EF = FB \times F A$
1 mark

Concepts used

trianglesparallelogramprovesimilar trianglespythagoras theorem

Approach

Step-by-step working

Concepts used

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