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In the diagram below, EO bisects side AC of $ riangle ACE$ - NSC Mathematics - Question 9 - 2016 - Paper 2
Question 9
In the diagram below, EO bisects side AC of $ riangle ACE$. EDO is produced to B such that BO = OD. AD and CD produced meet EC and EA at G and F respectively. 9.1 G... show full transcript
Worked Solution & Example Answer:In the diagram below, EO bisects side AC of $ riangle ACE$ - NSC Mathematics - Question 9 - 2016 - Paper 2
Step 1
Step 2
Write down, with reasons, TWO ratios each equal to ED/DB.
Answer
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From the property of and the side-splitter theorem, we have that ( \frac{ED}{FE} = \frac{ED}{DB} ) since segments DE and EF are parts of a larger triangle divided by a transversal.
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Also, from the same triangle properties, ( \frac{DB}{CG} = \frac{DB}{ED} ) holds as AD and CD are proportional segments divided by the equal height from point E to both segments.
Step 3
Prove that \( \hat{A_1} = \hat{F_2} \).
Answer
To prove that ( \hat{A_1} = \hat{F_2} ), we can use the property of alternate angles. Since line EF (extended) is parallel to line AB, and line AE acts as the transversal, we can conclude that ( \hat{A_1} = \hat{F_2} ) by the Alternate Interior Angles Theorem.
Step 4
Prove that ACGE is a cyclic quadrilateral.
Answer
Since ABCD is a rhombus, we know that the opposite angles are equal and adjacent angles are supplementary, leading to the conclusion that the angles subtended by the same chord AC in points G and E sum up to 180 degrees. Therefore, by the inscribed angle theorem, ACGE forms a cyclic quadrilateral.