Question

In the given figure, $AD\perp BC$ and $CD>BD$. Show that $AC>AB$.

Answer 1

Consider point $S$ on the line $BC$ so that $BD=SD$ and join $AS$.

Consider $△ADB$ and $△ADS$

We know that $SD=BD$

Since $AD$ is a perpendicular we know that

$\angle ADB=\angle ADS={90}^{\circ }$

$AD$ is common i.e. $AD=AD$

By $SAS$ congruence criterion

$△ADB\cong △ADS$

$AB=AS(c.p.c.t)$

Consider $△ABS$

We know that $AB=AS$

From the figure, we know that $\angle ASB$ and $\angle ABS$ are angles opposite to the equal sides

$\angle ASB=\angle ABS.\left(1\right)$

Consider $△ACS$

From the figure, we know that $\angle ASB$ and $\angle ACS$ are angles opposite to the equal sides

$\angle ASB=\angle ACS.\left(2\right)$

Considering the equations $\left(1\right)$ and $\left(2\right)$

$\angle ABS>\angle ACS$

It can be written as

$\angle ABC>\angle ACB$

So we get

$AC>AB$

Therefore, it is proved that $AC>AB$.