In ΔACB, ∠ACB=90° seg CD⊥side AB, A-D-B, Seg DE⊥side CB. Show that CD²×AC=AD×AB×DE. Best answers get Brainliest Rating.
Answers
Answer:
Given: ABC is a right triangle,
In which \angle ACB = 90^{\circ}∠ACB=90
∘
Also, CD \perp ABCD⊥AB and DE \perp ABDE⊥AB
Where D is any point in AB and E is any point in CB.
To prove : CD^2 \times AC= AD \times AB \times DECD
2
×AC=AD×AB×DE
Proof:
In triangles, ACB and ADC,
\angle CAB\cong \angle DAC∠CAB≅∠DAC ( Reflexive)
\angle ACB\cong \angle ADC∠ACB≅∠ADC ( Right angles)
Thus, By AA similarity postulate,
\triangle ACB\sim \triangle ADC△ACB∼△ADC -------(1)
⇒ \frac{AC}{AD} = \frac{AB}{AC}
AD
AC
=
AC
AB
⇒ AC^2= AB\times ADAC
2
=AB×AD -----------(2)
Now, Similarly,
\triangle ACB\sim \triangle CDB△ACB∼△CDB ----------(3)
Now, In triangles CED and CDB,
\angle ECD\cong \angle DCB∠ECD≅∠DCB ( Reflexive)
\angle CED\cong \angle CDB∠CED≅∠CDB ( Right angles)
By AA similarity postulate,
\triangle CED\sim \triangle CDB△CED∼△CDB ----------(4)
By equation (3) and (4),
\triangle ACB\sim \triangle CED△ACB∼△CED -------(5)
Again by equation (1) and (5)
\triangle ADC\sim \triangle CED△ADC∼△CED
\frac{AC}{CD} = \frac{DC}{ED}
CD
AC
=
ED
DC
CD^2 = AC\times DECD
2
=AC×DE ------------(6)
Multiplying equation (2) by CD^2CD
2
on both sides,
CD^2\times AC^2= CD^2\times AB\times ADCD
2
×AC
2
=CD
2
×AB×AD
From equation (6),
CD^2\times AC^2= AC\times DE \times AB\times ADCD
2
×AC
2
=AC×DE×AB×AD
CD^2\times AC= DE \times AB\times ADCD
2
×AC=DE×AB×AD
CD^2\times AC= AD\times AB\times DECD
2
×AC=AD×AB×DE
Hence Proved.
Answer:
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