In the given right triangle ABC right angled at B. BD is perpendicular to the hypotenuse AC. Then BD² =
1 point
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AD x BC
AB x CD
AB x BC
AD x CD
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In a right triangle ABC, the perpendicular BD on the hypotenuse AC is drawn. Prove that
(i)AC×AD=AB
2
(ii)AC×CD=BC
2
Medium
Solution
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Given: In a right-angled triangle △ABC,BD⊥AC.
To Prove: AC×AD=AB
2
AC×CD=BC
2
Proof: We draw a circle with BC as diameter. Since ∠BDC=90
∘
the circle on BC as diameter will pass through D.
Again, BC is a diameter and AB⊥BC. AB is tangent to the circle at B. Since AB is a tangent and AD is a secant to the circle.
⇒AC×AD=AB
2
Also,
AC×CD=AC×(AC−AD)
=AC
2
−AC×AD
=AC
2
−AB
2
[ Using AC×AD=AB
2
]
=BC
2
[△ABC is a right triangle]
Hence proved, AC×CD=BC
2
.
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SIMILAR QUESTIONS
star-struck
In △ ABC, AD is drawn perpendicular to BC:CD=3:1, then prove BC
2
=2(AB
2
−AC
2
).
Medium
View solution
>
In △ ABC, given that AB=AC and BD⊥AC. Prove that BC
2
=2AC⋅CD
Medium