Given: AB is perpendicular to BC, DC is perpendicular to BC, AC is congruent to DB Prove: AE is congruent to DE
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HEY------
---------DEAR---YOUR----'SOLUCTION---
Given: In ∆ABC and ∆DBE, AC⊥BC, BD ⊥BC and DE ⊥AB.
To prove BE/DE = AC/BC
Proof: In ∆ABC and ∆DBE,
‹c = ‹deb 90°
angle abc + angle DBE = 90°
Angle BDE + angle DBE = angle BD + Angle DBE
equal to angle abc equal to angle BDE
Triangle ABC congrance Triangle DBE
equal to AB by BD equal to AC by BE equal to DC by DE are common side
equal to AC by BC equal to BE by DE
Regards
MARK---
-----------AS----'BRAINLIEST
---------DEAR---YOUR----'SOLUCTION---
Given: In ∆ABC and ∆DBE, AC⊥BC, BD ⊥BC and DE ⊥AB.
To prove BE/DE = AC/BC
Proof: In ∆ABC and ∆DBE,
‹c = ‹deb 90°
angle abc + angle DBE = 90°
Angle BDE + angle DBE = angle BD + Angle DBE
equal to angle abc equal to angle BDE
Triangle ABC congrance Triangle DBE
equal to AB by BD equal to AC by BE equal to DC by DE are common side
equal to AC by BC equal to BE by DE
Regards
MARK---
-----------AS----'BRAINLIEST
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