In ΔABC, AD is perpendicular to BC and AC > AB. Prove that

(i) ∠CAD > ∠BAD

(ii) DC > BD

Given: ΔABC, AD is perpendicular to BC and AC > AB

To prove: (i) ∠CAD > ∠BAD

(ii) DC > BD

The figure to the given question is as shown below,

(i) Given AD is perpendicular to BC, so the ΔABC is divided into two right - angled triangles namely ΔABD and ΔADC.

Now consider ΔABD,

By Pythagoras theorem, we have

AB² = AD² + BD²……….(i)

Similarly in ΔADC,

by Pythagoras theorem, we have

AC² = AD² + DC²………(ii)

And it is given,

AC > AB

Squaring on both sides we get

AC² > AB²

Now substituting the values from equation (i) and (ii) in above equation, we get

AD² + DC² > AD² + BD²

Now cancelling the like terms on both sides we get

DC² > BD²

Taking square root on both sides, we get

DC > BD………(iii)

We know in a triangle the shortest side is always opposite the smallest interior angle and the longest side is always opposite the largest interior angle.

⇒∠CAD > ∠BAD

Hence proved

(ii) From equation (iii),

DC > BD

Hence proved