**1. ABCD is a sq.. M and N are the mid-points of CD and AD. BM and CN intersect at P. Prove that PM is perpendicular to CN.**=>Way of solution: If they're perpendicular each other, they form

**right angle**(s) at the point. Solution

**(1)**: focus on triangle PMC

**Outline**: triangle NDC and MCB are congruent (reason: S.A.S., not R.H.S.) 90degrees = angle MBC+angle BMC (prop. of sq; angle sum of triangle) = angle PMC + angle MCP (corr. angles, congruent triangles) angle PCM = 90 degrees also, by angle sum of triangle. Solution

**(2)**: focus on triangle PBC. prove the pair of congruent triangles, by corr. angles and prop. of sq., we know that angle PBC + angle PCB = 90 degrees. Thus, by exterrior angles of triangle, angle NPB is 90 degrees also. Solution

**(3)**: focus on polygon ANPB. Still, prove the pair of triangles and let angle PBC be x. By prop. of sq. and ext. angles of triangle, we know that angle ABP and angle ANP are 90-x and 90+x respectively, apply angle sum of polygon and get angle NPB is 90 degrees.

**Summary**: When we have to prove perpendicular lines,

**backward approch**must be useful --- and to find out where can we evaluate some right angles.

**(2)ABCD is a //gram; AE anf AF divides angle DAB into three equal parts, and BG bisects angle ABC. BG intersects AE anf AF at H and I respectively, find angle GIF + angle GHE.**Way to solve: sum of two angles without any length/angles given, it should be a "nice" number. Also, the angles are far from where information(bisect angle) is given, so ext. angle might useful. We Let angle GAI = y, by int. angle, angle ABH = 90 degrees-1.5y apply two times of ext. angles on triangle AGI and AGH and onbtain the two angles. The sum is 180 degrees. Alternative solution: Set up two variables, and you can obtain something like 3x+2z, which can be eliminate by equation from int. angles. summary: It is useful to set variables to shorten your steps.

**(7)In triangle ABC, D,E,F are mid-points of BC, CA and AB respectively, FE=EG, prove that E is the centroid od triangle ADG.**Way to solve: First ask yourself, how can we prove about centroid? You know that

**three medians intersects at one point**, so that proving two of them are medians OR prove one median being cut into ratio of 2:1 is enough. There are so many solution here, and only rough outline will be appeared here and not hard to get those result. For easier understanding, we "create" two reasons(tests) for centroid, (2 medians) and (median ratio 2:1) Let EC and DG intersect at I.

**Prove two medians:**By mid-point theorem, FE=BD=DC=EG, and FG//BC. Then, two approch makes the same affort.

**(1)**AEDF and DEGC are //gram (opp. sides eq.) then AH=HD and DI=IG. (prop. of //gram)

**(2)**triangle AFH and DEH congruent each other and triangle EGI and triangle DCI congruent each other, so that AH=HD and DI=IG (corr. sides, congruent triangles) Therefore, E is the centroid of triangle ADG (2 medians)

**Prove 2:1**: Apply mid-point thm. . By using the two method stated above, EI=IC=1.2(AE), and DI=IG (dorr. sides, congruent triangles) Therefore, E is the centroid of triangle ADG (median ratio 2:1) Another way to prove this is to prove HE:EApply mid-point thm. . By using the two method stated above, FH=HE=1.2(EG), and AH=HD (dorr. sides, congruent triangles) Therefore, E is the centroid of triangle ADG (median ratio 2:1) Summary: Proving something strange is not too hard, but you have to ask yourself the "test" about the provings. Here comes the Part two. http://allmaths.blogspot.com/2009/03/outline-of-solution-ch9-ws-1-part-two.html

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