Question

In the given figure, LM || CB and LN || CD. Prove that `(AM)/(AB) = (AN)/(AD)`.

Answer 1

Given: LM || CB and LN || CD in the figure C lies on LA, B on AM, D on AN.

To Prove: `(AM)/(AB) = (AN)/(AD)`.

Proof [Step-wise]:

1. Consider triangle AML. Point C lies on AL, and the line through C parallel to LM meets AM at B because CB || LM. By the Basic Proportionality Theorem (Thales), a line through a point on one side of a triangle parallel to another side divides the remaining side proportionally.

Hence, `(AB)/(BM) = (AC)/(CL)`.   ...(1)

2. Consider triangle ANL. Point C lies on AL, and the line through C parallel to LN meets AN at D because CD || LN.

By the same theorem, `(AD)/(DN) = (AC)/(CL)`   ...(2)

3. From (1) and (2) we get `(AB)/(BM) = (AD)/(DN)`.   ...(3)

4. Invert (3) to obtain `(BM)/(AB) = (DN)/(AD)`.   ...(4)

5. Compute `(AM)/(AB)` and `(AN)/(AD)` using AM = AB + BM and AN = AD + DN:

`(AM)/(AB) = (AB + BM)/(AB)`

= `1 + (BM)/(AB)` 

`(AN)/(AD) = (AD + DN)/(AD)` 

= `1 + (DN)/(AD)`

6. Using (4) `(BM)/(AB) = (DN)/(AD)` we get 

`(AM)/(AB) = 1 + (BM)/(AB)` 

= `1 + (DN)/(AD)` 

= `(AN)/(AD)`

Thus, `(AM)/(AB) = (AN)/(AD)`.