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10.1 In the diagram, \( \triangle ABC \) and \( \triangle ADEF \) are drawn such that \( \angle A = \angle D \), \angle B = \angle E \) and \( \angle C = \angle F \) - NSC Mathematics - Question 10 - 2022 - Paper 2
Question 10
10.1 In the diagram, \( \triangle ABC \) and \( \triangle ADEF \) are drawn such that \( \angle A = \angle D \), \angle B = \angle E \) and \( \angle C = \angle F \)... show full transcript
Worked Solution & Example Answer:10.1 In the diagram, \( \triangle ABC \) and \( \triangle ADEF \) are drawn such that \( \angle A = \angle D \), \angle B = \angle E \) and \( \angle C = \angle F \) - NSC Mathematics - Question 10 - 2022 - Paper 2
Step 1
Prove the theorem: if two triangles are equiangular then the corresponding sides are in proportion
Answer
To prove that ( \frac{AB}{AC} = \frac{DE}{DF} ), we will use the properties of similar triangles. Given that ( \angle A = \angle D ), \angle B = \angle E ), and ( \angle C = \angle F ), triangles ( ABC ) and ( ADEF ) are equiangular, hence by the AA criterion for similarity, we conclude that the two triangles are similar.
From the similarity of triangles, we can derive: [ \frac{AB}{AC} = \frac{DE}{DF} ] This establishes the required proportion as stated in the theorem.
Step 2
Prove, with reasons, that: FB || CG
Answer
To prove that ( FB \parallel CG ), we notice that ( BF \perp EC ) and ( EC ) is tangent to the circle at ( C ). By the tangent-secant theorem, this means that angles ( \angle FBC ) and ( \angle EGC ) are equal. Therefore, by the alternate interior angles theorem, since these angles are equal, we can conclude that ( FB \parallel CG ).
Step 3
Prove, with reasons, that: AFCB || ACDB
Answer
To prove that ( AFCB \parallel ACDB ), notice that angles ( \angle CFB ) and ( \angle CDA ) are both equal since they subtend the same arc ( AB ) in both triangles. By the Converse of the Alternate Angle Theorem, we can conclude that the lines are parallel, thus ( AFCB \parallel ACDB ).
Step 4
Step 5
Prove, with reasons, that CD² = CG · DB.
Answer
Using the property of circles, in ( \triangle CDB ), we apply the Power of a Point theorem which states that the product of the lengths from a point outside the circle to points on the circle is constant. Therefore, we have: [ CD^2 = CG \cdot DB. ]
Step 6
Hence, prove that DB = CG + FB.
Answer
From the earlier established relation ( CD^2 = CG \cdot DB ), we rearrange to express ( DB ) in terms of ( CG ) and observe that by geometry of the diagram and properties of the tangent segments, we have that: [ DB = CG + FB. ] This concludes the proof.