Question

if a perpendicular is drawn from the vertex of the right angle of a right triangle to hypotenuse then prove that the Triangles on the both side of the perpendicular are similar to the whole triangle and to each other​

Answer 1

ʜᴏᴘᴇ ʏᴏᴜ ɢᴏᴛ ᴜʀ ᴀɴsᴡᴇʀ.

❤ᴛʜᴀɴᴋ ʏᴏᴜ❤

Answer 2

Answer:

Step-by-step explanation:

Given :

ABC is right-angled triangle.

∠B=90

BD⊥AC

To prove :

(i) △ADB∼△BDC  

(ii) △ADB∼△ABC  

(iii) △BDC∼△ABC  

(iv) 2BD=AD×DC

(v) 2AB =AD×AC

(vi) 2BC=CD×AC

Proof:

(i) ∠ABD+∠DBC=90

Also, ∠C+∠DBC+∠BDC=180

∠C+∠DBC+90 =180

∠C+∠DBC=90

But, ∠ABD+∠DBC=90

∴ ∠ABD+∠DBC=∠C+∠DBC

⇒ ∠ABD=∠C ----- ( i )

Thus, in △ADB and △BDC,

⇒ ∠ABD=∠C [ From ( i ) ]

⇒ ∠ADB=∠BDC [ Each angle is 90  ]

∴ △ADB∼△BDC [ By AA similarity theorem ]

(ii)  In △ADB and △ABC

∠ADB=∠ABC [ Each angle 90  ]

∠A=∠A [ Common angle ]

∴ △ADB∼△ABC [ By AA similarity theorem ]

(iii)  In △BDC and △ABC,

⇒∠BDC=∠ABC [ Each angle is 90 ]

⇒∠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 ]

 2BD  =AD×DC

(v)  From (ii) we have,

△ADB∼△ABC

⇒AB AD  = AC AB  [ C.P.C.T. ]

∴ 2AB =AD×AC

(vi)  From (iii) we have,

△BDC∼△ABC

⇒AC BC = BC DC  [ C.P.C.T. ]

∴ 2BC =CD×AC