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secondary 4 | A Maths
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Jona
Jona

secondary 4 chevron_right A Maths chevron_right Singapore

Pls help. Thank you.

Date Posted: 4 years ago
Views: 200

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Eric Nicholas K
Eric Nicholas K's answer
5997 answers (Tutor Details)
1st
You need to first know about the concepts of congruency, similarity and the basic properties of circles and angles from E Maths before proceeding to solve these type of questions in A Maths.
Jona
Jona
4 years ago
Part 3 is not answered correctly.
Eric Nicholas K
Eric Nicholas K
4 years ago
I am sure this method can be used because I have done this type many times and this is the only method which I have used so far.

There is probably another method out there for use. Remember that proving questions can have more than one approach and of course, any valid approach is correct. If I can find another method I will write it down here.
Jona
Jona
4 years ago
FC & CE are not the full length of the 2 similar triangles, so how can we use the common ratio of the similar triangles to prove part iii?
Eric Nicholas K
Eric Nicholas K
4 years ago
Look again...

(You can extract out the two triangles concerned and redraw them on separate diagrams)

AFC and BEC are the two similar triangles concerned. There are six full lengths here, namely AF, AC, FC, BE, BC and EC. By the way, EC and CE are exactly the same length.

The fact that the triangles are similar means that all the corresponding angles have the same value and all the corresponding sides have equal ratios (including the case where their corresponding lengths are even equal).

Length AF corresponds to length BE. Length AC corresponds to length BC. Length FC corresponds to length EC. If you have formed the correct statement for the ratios of the lengths,

AF : BE = AC : BC = FC : EC or equivalently, AF/BE = AC/BC = FC/EC.

We only need two of the three terms in the equality, and in this case we pick AC/BC = FC/EC before cross-multiplying to obtain the solution. Remember that AC and so on are full lengths of the triangle.

If you do not believe that this is the case, you can always assign a value for the length of one side of triangle AFC and a length for the corresponding side of triangle BEC, following which a series of sine rules and cosine rules can be used to find the other lengths. From there, the relevant ratios can also be obtained.

This alternative method also works, but it’s much more tedious and there might be slight deviations in the values of the ratios due to some rounding errors.
Jona
Jona
4 years ago
Thank you very much for your explanation! Really appreciate it!
Just wondering, is it possible to prove that triangle ECF is similar to triangle ACB?
Eric Nicholas K
Eric Nicholas K
4 years ago
Based on the information provided in the equation, there is insufficient information required to prove that triangles ECF and ACB are equal. We are unable to prove that lines EF and AB are similar or the corresponding angles are equal.

The only other inference I can make out of this question is that A, B, E and F lie on a circle because the two labelled angles in the main question correspond to the angles in the same segment.
J
J
4 years ago
Actually it is possible to prove for ECF and ACB.


angle ECF = angle BCA (common angle)


Since CF x CB = CE x CA,

CF/CA = CE/CB

Since the included angle is the same and the ratios of the corresponding sides are equal,

by SAS , △ECF is similar to △ ACB

(The corresponding base angles that are equal are on opposite sides so you'll have to rotate either triangle.

i.e angle CEF and angle CBA are equal, and angle CFE and angle CAB are equal.