**Please don't copy the solution of no.4 bacause it's just a alternative way, without step and just for refrence.**====

**3. In the figure, ACDE is a sq. B is the mid point of CA, DE is produced to F such that CG=GF, and G is also lying on BE. GF and BE are stright lines. ai) Prove that triangle ABE congruent to EFA**Solution: Seems a stright forward question, but actually not. Note that BA =EF has NOT BEEN GIVEN yet. CA//DF can only give you a pair of (equal) right angle, which is stated also by prop. of sq. .BE // AF also has not been given yet. Therefore, the only thing you can do is prove BA=EF. Focus on triangle CBG, which is congruent to FEG, then CB=EF=BA (corr. sides; given), since you prove that, it's not hard at all, you'll got two sides plu one right angle, and the reason is S.A.S.

**aii) Prove triangle ACH similar to EFH.**Not hard. two times of alt. angles (or one time with prop. of sq.) plus a pair of vert. opp. angles give you A.A.A. for this.

**b) Hence, find area of triangle AHF: area of triangle of HEF**Note that both triangle have diff. base but the same height, so that this questions is same as AH:HE. Apply 2 sides prop. on the similar triangle, and got 2:1.

**Summary:**sometimes (a) is the most difficult part because something likely given (and true) hasn't been given to you. You should look carefully about the given information.

**4. ABCD is a //gram, with E is a (free) point lying on AB, EC and BD intersect at F., find area of triangle DEF : area of triangle CBF. (HKCEE 2007 MC)**Solution: Teacher said that it's all about primary knowledge, but I have no idea about the way to solve it. Let's take another way, just sensing and not a standard solution. Since E is a free point on AB, Let's let E is at the same position as A. By prop. of //gram, diag. bisect each other, and so that the two triangles are congruent! Well, 1:1 and this is a MC and don't too care the step. *xd*

**Summary:**When there's some free point in the figure, try to move these point and get solution, but never try to do this when steps are required.

**5.ABCD and ADEF is 2 identical rhombuses(shares the side AD), G is a point on AD such that BGE is a stright line. a)prove AG=GD.**Once you finish reading the question, mose probably triangle ADE will be concerned, but remember this' an rhombus, and "identical rhombuses" give you a pair of equal sides, and prop. of rhombus gives you two pair of equal angle by parallel lines, and triangle DGE is congruent to triangle AGE, by corr. sides, AG=DG.

**b)Hence, if AD is perpendicular to BE, prove AEF is an equilateral triangle.**In fact, "if" is rubbish because once you proved (a), AD is already perpendicular to BE (prop. of icos. triangle). You can find that triangle DGE is congruent to AGE, so that DE=AE(corr. sides), and it's not hard to get that AF=FE=DE=AE, so AEF is equilateral (by defination) Summary: Still, beware of the information that you had, never apply thing that you haven't got.

**6.ABFE, BCGF and CEHG are 3 identical sq.s, (sharing lines BF and CG) Let angle DEF = a, angle DFG=b, and angle DGH=c. Show that a+b+c is an right angle.**Just a verrrrry classical question. You will see this every year whatever in exam, competitions,... But let's follow the steps of worksheet.

**a)Show that triangle DFG is similar to triangle DEB.**A.A.A. is useless here, but there's lot of right angle here, which means Pyth. thm. is useful here. Let AB=x, and calculate those sides and apply 3 sides prop..

**b)show that angle BEF = a+b**Not hard. Apply corr. angles and prop. of sq. and got it.

**c)Show a+b+c = 90 degrees.**since angle BEF=angle DGH=45 degrees (prop. of sq.) their sum if 90 degrees.

**Warning**: DON'T think that ext. angle of triangle is useful here. Infinite prople has dumbly tried this and got a beautiful zero mark without the above steps. Ext. angle can only tell you pretty a lot of rubbish and return to the original problem for this type of question. .Summary: This is a claaical question of sum of degrees. You know nothing about the angles (c is 45, but a and b is irrational no.), and nothing can be applied. Maybe similar triangle can tell you some more pairs of equal angle and help you to find out the answer. That's all about the question, and welcome to ask me or provide more way to do it (esspecially no.4). Back to part one: http://allmaths.blogspot.com/2009/02/outline-of-solution-ch9-ws-1-part-one.html

## No comments:

## Post a Comment