Question

In triangle ABC, angle ACB =90 degree ,seg CD is perpendicular to side AB and seg DE is perpendicular to side CB . Show that square of CD × AC= AD × AB × DE

Answer 1

Answer:  

Given:  ABC is a right triangle,

In which $\angle ACB = 90^{\circ}$

Also, $CD \perp AB$ and  $DE \perp AB$

Where D is any point in AB and E is any point in CB.

To prove : $CD^2 \times AC= AD \times AB \times DE$

Proof:

In triangles, ACB and ADC,

$\angle CAB\cong \angle DAC$  ( Reflexive)

$\angle ACB\cong \angle ADC$   ( Right angles)

Thus, By AA similarity postulate,

$\triangle ACB\sim \triangle ADC$ -------(1)

⇒ $\frac{AC}{AD} = \frac{AB}{AC}$

⇒ $AC^2= AB\times AD$ -----------(2)

Now, Similarly,

$\triangle ACB\sim \triangle CDB$  ----------(3)

Now, In triangles CED and CDB,

$\angle ECD\cong \angle DCB$  ( Reflexive)

$\angle CED\cong \angle CDB$   ( Right angles)

By AA similarity postulate,

$\triangle CED\sim \triangle CDB$ ----------(4)

By equation (3) and (4),

$\triangle ACB\sim \triangle CED$ -------(5)

Again by equation (1) and (5)

$\triangle ADC\sim \triangle CED$

$\frac{AC}{CD} = \frac{DC}{ED}$

$CD^2 = AC\times DE$ ------------(6)

Multiplying equation (2) by $CD^2$ on both sides,

$CD^2\times AC^2= CD^2\times AB\times AD$

From equation (6),

$CD^2\times AC^2= AC\times DE \times AB\times AD$

$CD^2\times AC= DE \times AB\times AD$

$CD^2\times AC= AD\times AB\times DE$

Hence Proved.

Answer 2

Step-by-step explanation:

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 DECD2×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}ADAC=ACAB

⇒ AC^2= AB\times ADAC2=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}CDAC=EDDC

CD^2 = AC\times DECD2=AC×DE ------------(6)

Multiplying equation (2) by CD^2CD2 on both sides,

CD^2\times AC^2= CD^2\times AB\times ADCD2×AC2=CD2×AB×AD

From equation (6),

CD^2\times AC^2= AC\times DE \times AB\times ADCD2×AC2=AC×DE×AB×AD

CD^2\times AC= DE \times AB\times ADCD2×AC=DE×AB×AD

CD^2\times AC= AD\times AB\times DECD2×AC=AD×AB×DE

Ans : Hence CD² × AC = AD×AB × DE

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