Question

$△$ABC is a right angled triangle, right angled at B. BD is perpendicular to AC. What is AC . DC?

• A. BC.AB
• B. ${BC}^2$
• C. ${BD}^2$
• D. AB.AC

Answer 1

The correct option is B

${BC}^2$

Consider $△ADB$ and $△ABC$

∠BAD = ∠BAC [common angle]

∠BDA = ∠ABC [ ${90}^{\circ }$]

Therefore by AA similarity criterion, $△ADB$ and $△ABC$ are similar.

So, $\frac{AB}{AC}=\frac{AD}{AB}$ ⇒ ${AB}^2$ = AC.AD ---(I)

Similarly, $△BDC$ and $△ABC$ are similar.

So, $\frac{BC}{AC}$ = $\frac{DC}{BC}$ ⇒ ${BC}^2$ = AC.DC ---(II)

Dividing (I) and (II) and cancelling out AC, we get, ${\left(\frac{AB}{BC}\right)}^2$ = $\frac{AD}{DC}$ ----(III)

Also, In $△ADB$, ${AB}^2$ = ${AD}^2$ + ${DB}^2$ and in $△BDC$, ${CB}^2$ = ${CD}^2$ + ${DB}^2$ [Pythagoras theorem]

Subtracting these two equations above and cancelling off ${DB}^2$ on both sides, we get

${AB}^2$ - ${BC}^2$ = ${AD}^2$ - ${CD}^2$ ⇒ ${AB}^2$ + ${CD}^2$ = ${AD}^2$ + ${BC}^2$ --------------(IV)

Dividing this equation with ${BC}^2$ on both sides, $\frac{{AB}^2}{{BC}^2}$ + $\frac{{CD}^2}{{BC}^2}$ = $\frac{{AD}^2}{{BC}^2}$ + $\frac{{BC}^2}{{BC}^2}$ ⇒ $\frac{{AB}^2}{{BC}^2}$ + $\frac{{CD}^2}{{BC}^2}$ = $\frac{{AD}^2}{{BC}^2}$ + 1

$⟹$ $\frac{{AB}^2}{{BC}^2}$ - 1 = $\frac{{AD}^2}{{BC}^2}$ - $\frac{{CD}^2}{{BC}^2}$

$⟹$ $\frac{AD}{DC}$ - 1 = $\frac{{AD}^2-{CD}^2}{{BC}^2}$ [Using (III)]

$⟹$$\frac{AD-DC}{DC}$ = $\frac{(AD-CD)(AD+CD)}{{BC}^2}$

$⟹$$\frac{1}{DC}$ = $\frac{AC}{{BC}^2}$

$⟹$ ${BC}^2$ = AC.DC

Alternatively,

Consider $△ABC$ and $△BDC$,

$\angle ABC=\angle BDC={90}^{\circ }$

$\angle C=\angle C$ [Common angle]

Therefore by AA similarity criterion, $△ABC$ and $△BDC$ are similar.

$\frac{AC}{BC}$ = $\frac{BC}{DC}$

$B{C}^2$ = $AC\times DC$