The triangles are similar so all the corresponding angles will be equal.
The question does not state that the two lines BE and CD are parallel, so you cannot assume this.

It may help to draw the two triangles out separately next to each other.

4-8-edexcel-gcse-hard-q6

The angles BAE and CAD are the same angle so they are equal.

If the lines BE and CD are parallel, then the angle ABE = angle ACD then there will exist a value of k such that $\begin{array}{rcl} A C & = & k A B \end{array}$ , $\begin{array}{rcl} C D & = & k B E \end{array}$ and $\begin{array}{rcl} A D & = & k A E \end{array}$ .
k is known as the scale factor.

Form an equation using the two sides AE and AD.

$\begin{array}{rcl} A D & = & k A E \end{array}$

Substitute the values of $A E$ and $A D$ into the equation and solve to find $k$ .

$\begin{array}{rcl} 15 & = & 12 k \\ k & = & \frac{15}{12} = 1.25 \end{array}$

[1]

Substitute into $\begin{array}{rcl} A C & = & k A B \end{array}$ .

$\begin{array}{rcl} A C & = & k A B \\ 8 + x & = & (1.25) (8) \end{array}$

Solve to find $x$ .

$\begin{array}{rcl} 8 + x & = & 10 \\ x & = & 10 - 8 \\ x & = & 2 \end{array}$

[1]

If the lines BE and CD are not parallel, then it may instead be that angle ABE = angle ADC.
Drawing the triangles this way up instead will help.

4-8-edexcel-gcse-hard-q6-diagram2

In this case, there will exist a value of k such that $\begin{array}{rcl} A C & = & k A E \end{array}$ , $\begin{array}{rcl} C D & = & k B E \end{array}$ and $\begin{array}{rcl} A D & = & k A B \end{array}$ .
k is known as the scale factor.

Form an equation using the two sides AB and AD.

$\begin{array}{rcl} A D & = & k A B \end{array}$

Substitute the values of AB and AD into the equation and solve to find $k$ .

$\begin{array}{rcl} 15 & = & 8 k \\ k & = & \frac{15}{8} \end{array}$

Substitute into $\begin{array}{rcl} A C & = & k A E \end{array}$ .

$\begin{array}{rcl} A C & = & k A E \\ 8 + x & = & (\frac{15}{8}) (12) \end{array}$

[1]

Solve to find $x$ .

$\begin{array}{rcl} x + 8 & = & \frac{15 \times 12}{8} \\ x + 8 & = & \frac{15 \times 3}{2} \\ x & = & \frac{45}{2} - 8 \\ = & 22.5 - 8 \\ = & 14.5 \end{array}$

[1]

$x = 2$ cm, assuming that angle ABE = angle ACD
or $x = 14.5$ cm, assuming that angle ABE = angle ADC
Both assumptions given [1]

$\frac{R S}{S T}$	$\frac{Q S}{P T}$
$\frac{P T}{Q S}$	$\frac{R T}{R S}$

Congruence, Similarity & Geometrical Proof (AQA GCSE Maths)

Exam Questions

1. Number

Number Toolkit

Prime Factors, HCF & LCM

Powers, Roots & Standard Form

Fractions

Percentages

Simple & Compound Interest, Growth & Decay

Fractions, Decimals & Percentages

Rounding, Estimation & Bounds

Surds

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Factorising

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Algebraic Proof

Linear Equations

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Iteration

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Functions

Coordinate Geometry

Linear Graphs y = mx + c

Graphs of Functions

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