Question

In ΔAВC; D and E are two points on AB such that AD = DE = EB. Through D and E, lines are drawn parallel to AC meet BC at N and M, respectively. Through N and M, lines are drawn parallel to AB meet AC at Q and P. Prove that AP = PQ = CQ.

Answer 1

Given: In triangle ABC, points D and E lie on AB with AD = DE = EB.

Through D and E lines are drawn parallel to AC meeting BC at N and M respectively.

Through N and M lines are drawn parallel to AB meeting AC at Q and P respectively.

To Prove: AP = PQ = CQ.

Proof (Step-wise):

1. From AD = DE = EB we have

`AD = 1/3 AB` 

`AE = 2/3 AB`

`EB = 1/3 AB`

So, D divides AB at `1/3` from A, E at `2/3` from A.

2. Consider a line through D parallel to AC meeting BC at N.

Since DN || AC, triangles DBN and BAC are similar.

Hence, `(BN)/(BC) = (DB)/(BA)`.

But DB = AB – AD 

= ` 2/3 AB`

So, `(BN)/(BC) = 2/3`.

Therefore, CN = BC – BN

= `1/3 BC`

3. Similarly, the line through E parallel to AC meets BC at M.

From EM || AC we get

`(BM)/(BC) = (EB)/(BA) = 1/3`

So, `BM = 1/3 BC` and hence `CM = 2/3 BC`.

4. Now consider the line through N parallel to AB meeting AC at Q.

Because NQ || AB, triangles CNQ and CAB are similar.

So, `(CQ)/(CA) = (CN)/(CB)`. 

From step 2,

`(CN)/(CB) = 1/3`

Hence, `CQ = 1/3 CA`.

5. Likewise, the line through M parallel to AB meets AC at P.

Since MP || AB, triangles CMP and CAB are similar. 

So, `(CP)/(CA) = (CM)/(CB)`.

From step 3,

`(CM)/(CB) = 2/3`

Hence, `CP = 2/3 CA`.

Therefore, AP = CA − CP

= `CA - 2/3 CA`

= `1/3 CA`

6. Finally, on AC we have AP + PQ + QC = AC.

Substituting `AP = 1/3 CA` and `CQ = 1/3 CA`

Gives PQ = AC – AP – CQ 

= `CA - 1/3 CA - 1/3 CA`

= `1/3 CA`

7. Therefore, AP = PQ = CQ = `1/3` CA, as required.

AP = PQ = CQ; the points P and Q trisect AC.