Question

In the following figure, D is the mid-point of BC, and PQ || BC. Prove that AD bisects PQ.

Answer 1

Let AD intersect PQ at R,

Since PQ || BC, consider triangles △APR and △ABD.

We get:

• ∠APR = ∠ABD because AP lies on AB and PR || BD.
• ∠ARP = ∠ADB because AR lies on AD and RP || DB.

So, △APR ∼ △ABD,

Hence,

`(PR)/(BD) = (AR)/(AD)`

Now consider triangles △ARQ and △ADC.

Similarly,

• ∠AQR = ∠ACD
• ∠ARQ = ∠ADC

Therefore, △ARQ ∼ △ADC

Hence, `(RQ)/(DC) = (AR)/(AD)`

So, `(PR)/(BD) = (RQ)/(DC)`

But D is the midpoint of BC, therefore BD = DC

Hence, PR = RQ

Therefore, R is the midpoint of PQ,

So, AD bisects PQ.

Hence proved.