Question

In the following figure, BQ and CP are the altitudes on AC and AB, respectively. Prove that:

1. BP × OQ = CQ × OP
2. ΔРОQ − ΔВОС

Answer 1

Given that BQ and CP are altitudes of △ABC.

So,

• BQ ⊥ AC, hence ∠BQC = 90°
• CP ⊥ AB, hence ∠BPC = 90°

Also, O is the intersection point of BQ and CP.

⇒ Consider triangles △BOP and △COQ.

We have:

• ∠BPO = 90° and ∠CQO = 90°
• ∠BOP = ∠COQ since they are vertically opposite angles

Therefore, △BOP ∼ △COQ by AA similarity.

Hence, corresponding sides are proportional:

`(BP)/(CQ) = (OP)/(OQ)`

Cross-multiplying,

BP × OQ = CQ × OP

Hence proved,

⇒ In triangles △POQ and △BOC,

`(OP)/(OQ) = (BP)/(CQ)`

Now consider triangles △BPC and △CQB.

Since,

• ∠BPC = 90°
• ∠CQB = 90°
• ∠PBC = ∠QCB because PB lies on BA and QC lies on CA

So, △BPC ∼ △CQB

Therefore, ∠BCP = ∠CBQ, which gives the angle relation needed below.

Now in △POQ and △BOC:

• ∠POQ = ∠BOC (vertically opposite angles)
• ∠PQO = ∠BCO, since PQ lies along a direction corresponding to BC from the altitude construction, and equivalently, one can obtain this from the similarity of the right triangles formed by the altitudes.

Thus, △POQ ∼ △BOC by AA similarity.

Hence proved.