tc2019

the c Part spend me 1 hr !


pls help

ehbb

已知
1. 題目: BC = DE
2. 題目: <BCE = <CAD
3. (a)已證: <ADE = <ACB
4. (b)已證: 三角ABC跟三角AED同等
5. (c)提供: AD//BC

問題: AB x AE = BF x DE
其實貿貿然相乘我就真係唔識做
估計你資料(1)-(3)已用
而且問題涉及邊長考慮所以(1)-(3)未必太直接有用
(4)讓我聯想起比例
- AB = AE
- BC = ED
- CA = DA
問題: AB x AE = BF x DE
AB已有
AE已有
DE也有
未知數就係BF
而BF = BC-CF = ED-CF

未用資料: AD//BC
所以
- <AFB = <FAD = <BAC (因為同等三角)
- <DAC = <BCA
- <ADC+<ACF+<DCA = 180, 即2<ACF+<DCA = 180
- <ADC = <BCE

回到問題
仍有未知數CF
而要將邊長跟角度掛勾就只有相似三角
涉及CF既三角一係ACF一係就CEF
咁ACF應該冇邊個相似既
咁CFE呢? AFB幫唔幫到手?

試一試吧
<AFB = <CFE (vert. opp. <)
<ABF = <CEF (因為<ABC = <AED, 已證同等三角)
證據夠了嗎?
如是者CF邊長又可以同邊條邊長成比例呢?

ehbb

已知
1. 題目: BC = DE
2. 題目: <BCE = <CAD
3. (a)已證: <ADE = <ACB
4. (b)已證: 三角ABC跟三角AED同等
5. (c)提供: AD//BC

問題: AB x AE = BF x DE
其實貿貿然相乘我就真係唔識做
估計你資料(1)-(3)已用
而且問題涉及邊長考慮所以(1)-(3)未必太直接有用
(4)讓我聯想起比例
- AB = AE
- BC = ED
- CA = DA
問題: AB x AE = BF x DE
AB已有
AE已有
DE也有
未知數就係BF
而BF = BC-CF = ED-CF

未用資料: AD//BC
所以
- <AFB = <FAD = <BAC (因為同等三角)
- <DAC = <BCA
- <ADC+<ACF+<DCA = 180, 即2<ACF+<DCA = 180
- <ADC = <BCE

回到問題
仍有未知數CF
而要將邊長跟角度掛勾就只有相似三角
涉及CF既三角一係ACF一係就CEF
咁ACF應該冇邊個相似既
咁CFE呢? AFB幫唔幫到手?

試一試吧
<AFB = <CFE (vert. opp. <)
<ABF = <CEF (因為<ABC = <AED, 已證同等三角)
證據夠了嗎?
如是者CF邊長又可以同邊條邊長成比例呢?

qsdzzshawn

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yutszhin

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momolove

thanks for sharing