F.4 Maths trigonometry

Find a point D on BC such that AD丄BC. As cosA / a＝cosC / c,so, c cosA＝a cosCIn ΔABD, ΔACDBD＝c cosA ＝a cosC ＝CD∠ADB＝∠ADC＝90° ⋯⋯ (AD丄BC)AD＝AD ⋯⋯⋯⋯⋯⋯⋯⋯ (common)∴ ΔABD ≅ ΔACD ⋯⋯⋯⋯ (SAS)so, AB＝AC ⋯⋯⋯⋯⋯⋯ (corr. sides, ≅ Δ)∴ ΔABC is an isosceles triangle.

2015-06-11 08:59:00 補充：
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Find a point D on AC such that BD丄AC. As cosA / a＝cosC / c,
so, c cosA＝a cosC
In ΔABD, ΔCBD
AD＝c cosA
＝a cosC
＝CD
∠ADB＝∠CDB＝90° ⋯⋯ (BD丄AC)
BD＝BD ⋯⋯⋯⋯⋯⋯⋯⋯ (common)
∴ ΔABD ≅ ΔCBD ⋯⋯⋯⋯ (SAS)

2015-06-11 09:01:37 補充：
so, AD＝CD ⋯⋯⋯⋯⋯⋯ (corr. sides, ≅ Δ)
∴ ΔABC is an isosceles triangle.

2015-06-11 10:20:44 補充：
ABD is a right-angled triangle with ∠D＝90°, so
AD/AB = cos A
==> AD＝c cos A ⋯⋯ (AB＝c)
Similarly, CD＝a cos C
As c cos A＝a cos C
therefore, AD＝CD