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

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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 = -y°. Prove that AB/BD = AC/DC.

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

Step 1

Using the Sine Rule for Triangle ABD

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Answer

For triangle ABD, we can apply the sine rule. According to the sine rule:

ABsin(ADC)=BDsin(ABD)\frac{AB}{\sin(\angle ADC)} = \frac{BD}{\sin(\angle ABD)}

This can be rearranged to express AB/BD:

ABBD=sin(ADC)sin(ABD)\frac{AB}{BD} = \frac{\sin(\angle ADC)}{\sin(\angle ABD)}

Step 2

Using the Sine Rule for Triangle ACD

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Answer

Similarly, for triangle ACD, we again apply the sine rule:

ACsin(ABD)=DCsin(ADC)\frac{AC}{\sin(\angle ABD)} = \frac{DC}{\sin(\angle ADC)}

Rearranging this gives us:

ACDC=sin(ABD)sin(ADC)\frac{AC}{DC} = \frac{\sin(\angle ABD)}{\sin(\angle ADC)}

Step 3

Equating the Two Ratios

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Answer

From the expressions derived from the sine rule, we notice that:

ABBD=sin(ADC)sin(ABD)\frac{AB}{BD} = \frac{\sin(\angle ADC)}{\sin(\angle ABD)}

ACDC=sin(ABD)sin(ADC)\frac{AC}{DC} = \frac{\sin(\angle ABD)}{\sin(\angle ADC)}

By cross-multiplying the two ratios, we can affirm:

ABDC=ACBDAB * DC = AC * BD

Thus, we have shown that:

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

Step 4

Conclusion

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Answer

Therefore, we have proved the statement as required:

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

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