Question

In $△ABC,AP\perp BC,BQ\perp AC,B-P-C,A-Q-Cthenprovethat,△CPA\sim △CQB.IFAP=7,BQ=\&,BC=12thefindAC.$

Answer 1

To prove:$△CPA\cong △CQB$

Proof:In $△CPA$ and $△CQB$

$\angle CPA=\angle CQB={90}^{\circ }$ (given)

$\angle C=\angle C$ (common)

By AA similarity criterion,

$△CPA\cong △CQB$

Hence proved.

Now, $\frac{AP}{BQ}=\frac{AC}{BC}$ since corresponding sides are proportional.

$\Rightarrow AC=\frac{AP}{BQ}\times BC$

$=\frac{7}{8}\times 12=10.5$