Question

$△ABC$ is an isosceles triangle in which $AC=BC$. $AD$ and$BE$ are respectively two altitudes to sides $BC$ and $AC$. Prove that $AE=BD$.

Answer 1

Proving that, $\mathrm{AE}=\mathrm{BD}$

Given:

$△ABC$ is an isosceles triangle in which $\mathrm{AC}=\mathrm{BC}$.

To Prove:

$\mathrm{AE}=\mathrm{BD}$

Proof

According to given details

From$\Delta ADC$ and$\Delta BEC$,

We obtain

$\mathrm{AC}=\mathrm{BC}$ [Given]_______$\left(1\right)$

Since,$\angle \mathrm{ADC}=\angle \mathrm{BEC}=90°$

$\angle \mathrm{ADC}=\angle \mathrm{BEC}$ ($AD$ and $BE$ are altitude)

$\angle \mathrm{ACD}=\angle \mathrm{BCE}$ [Common angle]

So, $\mathrm{\Delta }\mathrm{ADC}\cong \mathrm{\Delta }\mathrm{BEC}$ [By $\mathrm{ASA}$congruence rule]

$\Rightarrow \mathrm{CE}=\mathrm{CD}$ [By CPCT]________$\left(2\right)$

Subtracting equation$\left(2\right)$ from$\left(1\right)$, We get

$\begin{array}{rcl}\mathrm{AC}–\mathrm{CE}& =& \mathrm{BC}–\mathrm{CD}\\ \mathrm{AE}& =& \mathrm{BD}\end{array}$

Hence, proved that $\mathrm{AE}=\mathrm{BD}$.