Question

please answer the question​

Answer 1

Answer:

Given triangle ABC, the altitude AP and BQ meet at O. we have to show that AO\times PO=BO\times QOAO×PO=BO×QO

In ΔAOQ and ΔBOP

∠OPB=∠OQA (∵each 90°)

∠BOP=∠AOQ (∵vertically opposite angles)

By AA similarity rule, ΔAOQ is similar to ΔBOP

∴ by theorem, sides are proportional as

\frac{BO}{AO}=\frac{PO}{QO}

AO

BO

=

QO

PO

⇒ AO\times PO=BO\times QOAO×PO=BO×QO

Hence, proved