演繹幾何三條（F.3 Maths）

1a. Since CA=CB and KA=KB (given)
also, CK=CK (common side)
thus ΔAKC congruent to ΔBKC (SSS)

Hence, ∠ACK=∠BCK (corr. ∠s, congruent Δs)
i.e. KH is the angle bisector of ∠AKB.

b. Since CA=CB (given), and ∠ACK=∠BCK (proven)
also we have CH=CH (common side)
therefore we have ΔAHC congruent to ΔBHC (SAS)

Now we have AH=BH (corr. sides, congruent Δs)
also ∠AHC=∠BHC (corr. ∠s, congruent Δs)

Moreover, we have ∠AHC+∠BHC=180 degree (adj. ∠s at st. line)
Therefore we have ∠AHC=∠BHC=90 degree.
i.e. CH⊥AB

Since we have AH=BH and CH⊥AB,
thus KH is perpendicular bisector of AB.

2a. Since ΔADE congruent to ΔCBF (given)
thus DE=BF and AE=CF (corr. sides, congruent Δs)

Now since EF=FE (common side)
thus AF=AE+EF=CF+FE=CE.

On the other hand, we have ∠AED=∠CFB (corr. ∠s, congruent Δs)
also, ∠AED+∠DEC=∠CFB+∠BFA=180 degree (adj. ∠s at st. line)
therefore, ∠DEC=∠BFA.

Hence, ΔCDE congruent to ΔABF (SAS)

b1. Since ∠BAF+∠CDE=90 degree (given)
but as we have proven that ΔCDE congruent to ΔABF,
we have ∠BAF=∠DCE (corr. ∠s, congruent Δs)
thus ∠CDE+∠DCE=90 degree.

Moreover, since ∠CDE+∠DCE+∠DEC=180 degree (∠ sum of Δ)
thus we have ∠DEC=90 degree.
i.e. DE is an altitude of ΔACD.

b2. Similar to (b1).

Since ∠BAF+∠CDE=90 degree (given)
but as we have proven that ΔCDE congruent to ΔABF,
we have ∠CDE=∠ABF (corr. ∠s, congruent Δs)
thus ∠ABF+∠BAF=90 degree.

Moreover, since ∠ABF+∠BAF+∠AFB=180 degree (∠ sum of Δ)
thus we have ∠AFB=90 degree.
i.e. BF is an altitude of ΔABC.

2008-12-27 22:40:43 補充：
3a. Construct the altitude of BC in ΔABC, say AH.
then
Area of ΔABD=1/2*BD*AH
=1/2*(1/2*BC)*AH..............(Since AD is median, BD=1/2*BC)
=1/2*Area of ΔABC.

Thus
Area of ΔABD / Area of ΔABC = 1 / 2

2008-12-27 22:40:52 補充：
b. Since Area of ΔAKE = Area of ΔBKD (given)
Area of ΔAKE+Area of ΔAKB = Area of ΔBKD+Area of ΔBKA
Area of ΔABD = Area of ΔABE
1/2*Area of ΔABC = Area of ΔABE

By the same arguent in (a), we know that AE=1/2*AC
hence BE is a median of ΔABC.

2008-12-27 22:41:48 補充：
唔好意思，因為一次過答晒超過左字數限制，
所以分開左3 parts，希望睇得明啦！=]