ASSIGNMENT - 4 Marks
In a right-angled triangle ABC, the perpendicular drawn
from the point C on the hypotenuse meets the
hypotenuse at D and the bisector of the 2C meets
AE
the hypotenuse at E. Prove that
DB
Answers
AD/BD = AE²/BE²
Step-by-step explanation:
to be proved
AD/DB=AE²/BE²
CE is angle bisector of ∠C
=> AE/BE = AC/BC
CD ⊥ AB
=> ΔADC ≈ ΔCDB
=> AD/CD = AC/BC = CD/BD
=> AD * BD = CD²
AC² = AD² + CD² = AD² + 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/BE
=> AE²/BE² = AD/BD
=> AD/BD = AE²/BE²
QED
Proved
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Answer:
Step-by-step explanation:
to be proved
AD/DB=AE²/BE²
CE is angle bisector of ∠C
=> AE/BE = AC/BC
CD ⊥ AB
=> ΔADC ≈ ΔCDB
=> AD/CD = AC/BC = CD/BD
=> AD * BD = CD²
AC² = AD² + CD² = AD² + 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/BE
=> AE²/BE² = AD/BD
=> AD/BD = AE²/BE²