Question

$ABC$ is an isosceles triangle in which $AC=BCAD$ and $BE$ are respectively two altitudes of side $BC$ and $AC$. Prove that $AE=BD$.

Answer 1

In $△ABC,AB=AC.$Also as it is isosecles triangle So based angle

$\angle ABC=\angle ACB.AD$ and $BE$ are the altitude on sides $BC$ and $AC$ respectively

In $△ABD$ and $△ACD,$

$\left(1\right)AD$ is common .

$\left(2\right)\angle ADB=\angle ADC={90}^o$

$\left(3\right)AB=AC$(given)

$\therefore △ABD\cong △ACD$(SAS congruency)

$\therefore \angle BAD=\angle DAC$ and $BD=DC\left(1\right)$

In $△ABD$ and $△ABE,$

$\left(1\right)AB$ is common

$\left(2\right)\angle ADB=\angle AEB={90}^o$

$\left(3\right)\angle BAD=\angle DAE$ (proved in 1)

$\therefore △ABD\cong △DAE$ (SAA congruency)

$\therefore BD=DE$proved