Question

In the given figure, AD ⊥ 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

∠ ADB = ∠ ADS = 90^o

AD is common i.e. AD = AD

By SAS congruence criterion

△ ADB ≅ △ ADS

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

Consider △ ABS

We know that AB = AS

From the figure we know that ∠ ASB and ∠ ABS are angles opposite to the equal sides

∠ ASB = ∠ ABS …. (1)

Consider △ ACS

From the figure we know that ∠ ASB and ∠ ACS are angles opposite to the equal sides

∠ ASB = ∠ ACS …. (2)

Considering the equations (1) and (2)

∠ ABS > ∠ ACS

It can be written as

∠ ABC > ∠ ACB

So we get

AC > AB

Therefore, it is proved that AC > AB.