in a right angled triangle ABC,the perpendicular is drawn from the point c on the hypotenusat D,and the bisectorof angle C meets the hypotnuse at E prove that AD÷DB=AEsquare ÷ EV square
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AD/DB = AE²/EB²
Step-by-step explanation:
CE is angle bisector of ∠C
=> AE/EB = AC/BC
CD ⊥ AB
=> ΔADC ≈ ΔCDB
=> AD/CD = AR/CB = DC/DB
=> AD * DB = DC²
AC² = AD² + CD² = BD² + AD * BD = AD ( AD + BD)
BC² = BD² + CD² = BD² + AD * BD= BD ( BD + AD)
AC²/BC² = AD ( AD + BD) / BD ( BD + AD)
=> AC²/BC² = AD/BD
AC/BC = AE/EB
=> AE²/EB² = AD/BD
=> AD/DB = AE²/EB²
QED
Proved
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