Question

plz help me in this maths question of geometry ( similarity) hope you will help me to complete this homework and I will mark you as brainliest​

Answer 1

Answer:

In ∆ALM and ∆ACB

<A=<A (COMMON)

<ALM=<ACB(LM II CB)

THEREFORE, ∆ALM~∆ACB

THEREFORE, AM/AB=AL/AC(CPST) ............ 1

IN ∆ANL and ∆ADC

<A=<A(COMMON)

<ALN=<ACD(LN II CD)

THEREFORE, ∆ANL~∆ADC

THEREFORE, AN/AD=AL/AC(CPST)..............2

From 1&2

AM/AB=AN/AD

Hence, proved