In ABC- - ACB -90- CD - AB- Prove thatBC 2- AC 2- BD - AD

Δ ADC and Δ ACB ∠ ADC = ∠ ACB = 90^∘ ∠ CAD = ∠ CAB (common) So, Δ ADC ∼Δ ACB by AA AD/AC=AC/AB AC^2 = AD . AB …. (i) Δ BCD and Δ BAC ∠ CAD = ∠ CBA (commo ...

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Standard X

Mathematics

AAA Similarity

In ABC, ∠ ACB...

Question

In $△$ ABC, $∠ A C B = 90^{\circ}, C D ⊥ A B$ . Prove that
$\frac{B C^{2}}{A C^{2}} = \frac{B D}{A D}$

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Solution

$Δ A D C$ and $Δ A C B$
$∠ A D C$ = $∠ A C B = 90^{\circ}$
$∠ C A D = ∠ C A B$ (common)
So, $Δ A D C \sim Δ A C B$ by AA
$\frac{A D}{A C} = \frac{A C}{A B}$
$A C^{2} = A D . A B$ …. (i)
$Δ B C D a n d Δ B A C$
$∠ C A D = ∠ C B A$ (common)
$∠ B D C = ∠ B C A = 90^{\circ}$
So, $Δ B C D \sim Δ B A C$
By AA similarity
$\frac{B C}{B A} = \frac{B D}{B C}$
$B C^{2} = B A \times B D$ ...(ii)
From (i) and (ii)
$\frac{A C^{2}}{B C^{2}} = \frac{A D}{B C}$ (Proved)

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In △ ABC, ∠ACB=90∘,CD⊥AB. Prove that BC2AC2=BDAD

ΔADC and ΔACB ∠ADC = ∠ACB=90∘ ∠CAD=∠CAB (common) So, ΔADC∼ΔACB by AA ADAC=ACAB AC2=AD.AB …. (i) ΔBCDandΔBAC ∠CAD=∠CBA (common) ∠BDC=∠BCA=90∘ So, ΔBCD∼ΔBAC By AA similarity BCBA=BDBC BC2=BA×BD...(ii) From (i) and (ii) AC2BC2=ADBC (Proved)

In $△$ ABC, $∠ A C B = 90^{\circ}, C D ⊥ A B$ . Prove that
$\frac{B C^{2}}{A C^{2}} = \frac{B D}{A D}$