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ABC is a triangle - Edexcel - GCSE Maths - Question 1 - 2021 - Paper 3

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Question 1

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ABC is a triangle. D is the point on BC such that angle BAD = angle DAC = - x° Prove that AB/BD = AC/DC (Total for Question 23 is 4 marks)

Worked Solution & Example Answer:ABC is a triangle - Edexcel - GCSE Maths - Question 1 - 2021 - Paper 3

Step 1

Express the Areas of Triangles ABD and ACD

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Answer

To prove that ( \frac{AB}{BD} = \frac{AC}{DC} ), we start by considering the areas of triangles ABD and ACD.

Using the sine rule, we can express the areas as follows:

The area of triangle ABD can be calculated as: AreaABD=12×AB×BD×sin(ADB)Area_{ABD} = \frac{1}{2} \times AB \times BD \times \sin(\angle ADB)

Similarly, the area of triangle ACD is: AreaACD=12×AC×DC×sin(ADC)Area_{ACD} = \frac{1}{2} \times AC \times DC \times \sin(\angle ADC)

Given that ( \angle ADB = \angle ADC ), we can simplify our expressions for both areas.

Step 2

Apply the Sine Rule

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Answer

From the sine rule, we can say:

ABAC=BDsin(ACD)DCsin(ABD)\frac{AB}{AC} = \frac{BD \cdot \sin(\angle ACD)}{DC \cdot \sin(\angle ABD)}

Since ( \angle ABD = -x° ) and ( \angle ACD = -x° ), we find that:

sin(ABD)=sin(ACD)\sin(\angle ABD) = \sin(\angle ACD)

Therefore, we have: ABAC=BDDC\frac{AB}{AC} = \frac{BD}{DC}

Step 3

Derive the Required Proportion

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Answer

This relationship can be rearranged to yield:

ABBD=ACDC\frac{AB}{BD} = \frac{AC}{DC}

which is the required proof as stated in the problem. Thus we have shown that the ratios of the sides are equal.

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