Given: ∠BDC=90 ∘ ∠ABC=90 ∘ Let ∠BCD=x ∘ Using triangle sum property in △BDC, ∠DBC=90−x ∘ Also ∠ABD=x ∘ Using triangle sum property in △ADB, ∠BAD=90−x ∘ Now considering △BDC and △ADB ∠BDC=∠BDA [∵∠BDC=∠BDA=90 ∘ ] ∠DBC=∠DAB [∵∠DBC=∠DAB=(90−x) ∘ ] So by AA △BDC∼△ADB Hence CD BD = BD AD CD 8 = 8 4 CD=16 Hence CD=16cm Reply
Given:
∠BDC=90
∘
∠ABC=90
∘
Let ∠BCD=x
∘
Using triangle sum property in △BDC, ∠DBC=90−x
∘
Also ∠ABD=x
∘
Using triangle sum property in △ADB, ∠BAD=90−x
∘
Now considering △BDC and △ADB
∠BDC=∠BDA [∵∠BDC=∠BDA=90
∘
]
∠DBC=∠DAB [∵∠DBC=∠DAB=(90−x)
∘
]
So by AA
△BDC∼△ADB
Hence
CD
BD
=
BD
AD
CD
8
=
8
4
CD=16
Hence CD=16cm