Question

class 10 ex 6.2 please give me a answer hurry......​

Answer 1

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

In ∆ACB , LM //CB and

In ∆ADC , NL // DC.

$\underline{\blue{ RTP : }}$

$\red{ \frac{AM}{AB} = \frac{AN}{AD} }$

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

$In \: \triangle ACB , LM \parallel CB \\\implies \frac{AM}{AB} = \frac{AL}{AC} \: --(1)$

$\blue{ \because B.P. Theorem ) }$

$In \: \triangle ADC , NL\parallel DC \\\implies \frac{AN}{AD} = \frac{AL}{AC} \: --(2)$

$\blue{ \because B.P. Theorem ) }$

$From \: (1) \:and \: (2)$

$we \: get$

$\green{ \frac{AM}{AB} = \frac{AN}{AD} }$

•••♪