數學:用角度證明AC=BC?

Sol
用角度證明
∠BAD=90度>∠ABD
過A作一直線文BD線段為E使得∠BAE=∠ABD
AE=BE
∠DAE=90度－∠BAE=90度－∠ABD=∠ADB
∠DAE==∠ADB
AE=DE
BE=DE
E為BD中點

C=E
AE=BE=ED
AC=BC=CD
AC=BC
△ABC為等腰△

BC / sin∠BAC = AC / sin∠ABC
AC / BC = sin∠ABC / sin ∠BAC

CD / sin∠CAD = AC / sin∠ADC
AC / CD = sin∠ADC / sin∠CAD
AC / BC = cos∠ABC / cos∠BAC

sin∠ABC / sin∠BAC = cos∠ABC / cos∠BAC
tan∠BAC = tan∠ABC
∠BAC = ∠ABC
∴ AC = BC　　( 等角對邊相等 )
∴ AC = BC = CD

∵ AC = BC
∴ ΔABC 為等腰Δ

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∵ ∠BAD = 90°
∴ BD 為直徑　　( 半圓上的圓周角的逆定理 )
BC = CD，則 C 為圓心
AC 為半徑
∴ AC = BC = CD

∵ AC = BC
∴ ΔABC 為等腰Δ