Question

D is a point on the side BC of a triangle
ABC such that angle ADC = angle BAC. Show
that CA^2=CB.CD
please solve quickly i promise mark your answer as brainliest​

Answer 1

Answer:

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• ∠ADC = ∠BAC

$\Large{\underline{\underline{\bf{To\:Prove:}}}}$

• CA² = CB × CD

$\Large{\underline{\underline{\bf{Proof:}}}}$

→ Consider Δ ABC and ΔDAC

  ∠BAC = ∠DAC ( given)

  ∠C = ∠C  ( common)

→ Hence,

  ΔABC $\sim$ Δ DAC by AA criteria.

→ Therefore,

  $\dfrac{AB}{AD} = \dfrac{BC}{AC} =\dfrac{AC}{DC}$

→ Consider the second part of the equation

  $\dfrac{BC}{AC}=\dfrac{AC}{DC}$

→ Cross multiplying

 AC × AC = BC × DC

 AC² = BC × DC

→ Hence proved.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Two triangles are similar if the ratio of their corresponding sides are equal.(SSS)

→ Two triangles are similar if two of their angles are equal.(AA)

→ Two triangles are similar if ratio of two of their sides are equal and the included angle is also equal ( SAS )