in any right angle triangle if a perpendicular is drawn from the right angle point on the hypothesis the two triangle on both and each of them is similar to the right angle ?
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Answers
Answer:
Given :
Here, ABC is a right-angled triangle.
∠B=90
o
and BD⊥AC
To prove :
(i) △ADB∼△BDC
(ii) △ADB∼△ABC
(iii) △BDC∼△ABC
(iv) BD
2
=AD×DC
(v) AB
2
=AD×AC
(vi) BC
2
=CD×AC
Proof:
(i)
⇒ ∠ABD+∠DBC=90
o
⇒ Also, ∠C+∠DBC+∠BDC=180
o
⇒ ∠C+∠DBC+90
o
=180
o
⇒ ∠C+∠DBC=90
o
But, ∠ABD+∠DBC=90
o
∴ ∠ABD+∠DBC=∠C+∠DBC
⇒ ∠ABD=∠C ----- ( 1 )
Thus, in △ADB and △BDC,
⇒ ∠ABD=∠C [ From ( 1 ) ]
⇒ ∠ADB=∠BDC [ Each angle is 90
O
]
∴ △ADB∼△BDC [ By AA similarity theorem ]
(ii)
In △ADB and △ABC
⇒ ∠ADB=∠ABC [ Each angle 90
o
]
⇒ ∠A=∠A [ Common angle ]
∴ △ADB∼△ABC [ By AA similarity theorem ]
(iii)
In △BDC and △ABC,
⇒ ∠BDC=∠ABC [ Each angle is 90
o
]
⇒ ∠C=∠C [ Common angle ]
∴ △BDC∼△ABC [ By AA similarity theorem ]
(iv)
From (i) we have,
△ADB∼△BDC
⇒
BD
AD
=
DC
BD
[ C.P.C.T ]
⇒ BD
2
=AD×DC
(v)
From (ii) we have,
△ADB∼△ABC
⇒
AB
AD
=
AC
AB
[ C.P.C.T. ]
∴ AB
2
=AD×AC
(vi)
From (iii) we have,
△BDC∼△ABC
⇒
AC
BC
=
BC
DC
[ C.P.C.T. ]
∴ BC
2
=CD×AC
Step-by-step explanation:
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