Question

In ΔABC, BP and CQ are altitudes from B and C on AC and AB respectively. BP and CQ intersect at O. Prove that
(i) PC x OQ = QB x OP

(ii) `"OC"^2/"OB"^2 = ("PC" xx "PO")/("QB" xx "QO")`

Answer 1

(i) In ΔOBQ and ΔOPC
∠OQB = ∠OPC = 90°  ...(QC and BP are altitudes)
∠QOB = ∠POC            ...(vertically opposite angles)
Therefore, ΔOBQ ∼ ΔOPC

⇒ `"PC"/"OP" = "QB"/"OQ"`
⇒ PC x OQ = QB x OP

(ii) Since ΔOBQ ∼ ΔOPC

`"OC"/"PO" xx "OC"/"PC" = "OB"/"QB" xx "OB"/"QO"`

⇒ `"OC"^2/("PC" xx "PO")  = "B"^2/("QB" xx QO")`

⇒ `"OC"^2/"OB"^2 = ("PC" xx "PQ")/("QB" xx "QO")`.