Question

D is the point on the side BC of triangle ABC such that angle ADC is equal to angle BAC. Prove that : CA:(square:) = CB×CD. ​

Answer 1

⋆ $\bold{\underline{\underline{\rm{\red{GivEn:-}}}}}$

• ∆ABC where ∠ADC = ∠BAC

⋆ $\bold{\underline{\underline{\rm{\blue{To\;ProvE:-}}}}}$

• CA² = CB×CD.

⋆ $\bold{\underline{\underline{\rm{\purple{Proof:-}}}}}$

★ In ∆ BAC and ∆ ADC

∠ACB = ∠ADC ....$\sf{ \bigg[ ∵ Common\;Angle \bigg]}$

∠BAC = ∠ADC .....$\sf{ \bigg[ ∵ Given \bigg]}$

∴ Hence, By AA similarity criterion

∆ BAC ∼ ∆ADC

Therefore,

$\sf{ \dfrac{BA}{AD} = \dfrac{AC}{CD} = \dfrac{BC}{AC}}$

$\sf{ \bigg[ ∵ Since \; two \; triangle\; are \; similar\; thier\; corresponding \;sides \;are \;proportional. \bigg]}$

∴ Hence,

$\sf{ \dfrac{AC}{CD} = \dfrac{BC}{AC}}$

★ Cross multiplication:-

AC × AC = BC × CD

AC² = BC × CD

Hence Proved!

__________________

Answer 2

${\huge{\bold{\blue{\bigstar{\boxed{\bf{solution}}}}}}}$

${\huge{\underline{\pink{GIVEN}}}}$

• $∠ADC = ∠BAC$

${\huge{\underline{\pink{To \:prove}}}}$

• $CA^2 = CB \times CD$

${\huge{\underline{\sf{\pink{Proof}}}}}$

In ΔADC and ΔBAC

∠ADC = ∠BAC (Given)

∠ACD = ∠BCA (Common angle)

∴ ΔADC ~ ΔBAC (By AA similarity criterion)

• when the triangles are similar their corresponding sides are in proportion

${\bold{\blue{\boxed{\bf\dfrac{BA}{AD}= \dfrac{AC}{CD} = \dfrac{BC}{AC}}}}}$

therefore ;

$\sf{\dfrac{AC}{CD}} = {\dfrac{BC}{AC}}$

• cross multiplication

$AC\times AC = CD\times BC$

................Hence proved

_____________________❤