Question

ABC is a right angled triangle at A. AD is perpendicular from A on BC.
1.Prove that ABDA~AADC
ii.Calculate BC,if AD =3cm and BD=2cm.​

Answer 1

pick 1st is answer of question no 1st and pick 2nd is of question number 2nd

Answer 2

$\underline { \blue { Given :}}$

$A \:right \: triangle \:ABC \: right \:angled\\ at\:A. AD \perp BC.$

$\underline { \blue { To \:prove :}}$

$i) \triangle BDA \sim \triangle ADC.\\ii) BC = ?$

$\underline { \green { Proof :}}$

$i) We \:have , \\\angle {BAD} + \angle {DAC} = 90\degree \\\angle C + \angle {DAC} + \angle {ADC} = 180\degree \\\blue {( Angle \:sum \: property )}$

$\implies \angle C + \angle {DAC} + 90= 180\degree$

$\implies C + \angle {DAC} = 90\degree$

$But \: \angle {BAD} + \angle {DAC} = 90\degree$

$\angle {BAD} + \angle {DAC} = \angle C + \angle {DAC}$

$\implies \angle {BAD} = \angle C \: --(1)$

$In \:\triangle BDA \:and \:\triangle ADC , \\we \:have ,\\\angle {BAD} = \angle C \: [From \:(1) ]$

$and \: \angle {BDA} = \angle {ADC} \\\blue {( Each \:equal \:to \: 90\degree) }$

$\triangle BDA \sim \triangle ADC \\\blue { (By \: A.A.\: Similarity \: Criterion )}$

$ii) From \:(1) , we \:have , \\\triangle BDA \sim \triangle ADC\\\implies \frac{BD}{AD} = \frac{AD}{DC} \\\implies \frac{2}{3} = \frac{3}{CD}$

$\implies CD = \frac{9}{2} = 4.5 \:cm$

$Now, BC = CD + DB \\= 4.5 \:cm + 2 \:cm \\= 6.5 \:cm$

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