Question

In $△ABC,AB=AC,andCPandBQ$ are altitudes. Prove that $CP=BQ$.

Answer 1

REF.Image

In $△ABC$,

$AB=AC$

Since CP & BQ are

altitudes, so :

$\angle CPB=\angle BQC={90}^{\circ }$ ___ (1)

and side AB = AC

and as angle opposite to

equal sides are also equal so

$\angle ABC=\angle ACB$ ___ (2)

$\to$ In $△CPB$ & $△BQC$;

$\angle BPC=\angle BQC={90}^{\circ }$ (from (1))

$\angle PBC=\angle QCB$ (from (2))

and BC = BC (common)

so $△CPB\cong △BQC$ (By ASA congruency rule)

$\therefore$ By CPCT, We can sat that BQ = PC

$\therefore BQ=PC$

Hence proved.