in a triangle ABC, angle ACB equal to 90 degree and CD perpendicular to AB prove that BC square by AC square is equal to BD by AD
Answers
Answered by
0
BC²/AC² = BC/AD ,
Step-by-step explanation:
ΔACB & ΔADC
∠A = ∠A ( Common)
∠ACB = ∠ADC = 90°
=> ΔACB ≈ Δ ADC
=> AC/AD = BC/CD
=> BC/AC = CD/AD
ΔBCA & ΔBDC
∠B = ∠B ( Common)
∠BCA = ∠BDC = 90°
=> ΔBCA ≈ Δ BDC
=> BC/BD = AC/CD
=> BC/AC = BD/CD
(BC/AC) * (BC/AC) = (CD/AD) * (BD/CD)
=> BC²/AC² = BC/AD
QED
Proved
Learn more:
Right triangle ABC is shown. Triangle ABC is shown. Angle ACB is a ...
https://brainly.in/question/14667895
in figure 7.33 BD and CE are altitudes of triangle ABC such that BD ...
https://brainly.in/question/13399135
Answered by
4
Answer:
See the attachment. Hope it helps uh!
Attachments:
Similar questions