Question

In given figure if a perpendicular is drawn from the right angle vertex of a right triangle to the hypotenuse, then prove that the triangle on each side of the perpendicular is similar to each other and to the original triangle. Also, prove that the square of the perpendicular is equal to the product of the lengths of the two parts of the hypotenuse.

Answer 1

Given a right triangle $ABC$ right angled at $B$ and $BD\perp AC$

To prove

(i) $△ADB\sim △BDC$

(ii) $△ADB\sim △ABC$

(iii) $△BDC\sim △ABC$

(iv) ${BD}^2=AD\times DC$

Solution:

(i) In $△ADB$ and $△BDC$, we have:

$\angle ABD+\angle DBC={90}^0$

Also $\angle C+\angle DBC+\angle BDC={180}^0$

$\Rightarrow$ $\angle C+\angle DBC+{90}^0={180}^0$

$\Rightarrow$ $\angle C+\angle DBC={90}^0$

But, $\angle ABD+\angle DBC={90}^0$

$\therefore$ $\angle ABD+\angle DBC=\angle C+\angle DBC$

$\Rightarrow$ $\angle ABD=\angle C$.........(1)

Thus, in $△ADB$ and $△BDC$, we have:

$\angle ABD=\angle C$ [From (1)]

and, $\angle ADB=\angle BDC$ [Each equal to ${90}^0$]

So, by AA-similarity criterion, $△ADB\sim △BDC$ $\left[Henceproved\right]$

(ii) In $△ADB$ and $△ABC$, we have:

$\angle ADB=\angle ABC$ [Each equal to ${90}^0$]

and, $\angle A=\angle A$ [Common]

So, by AA-similarity criterion, $△ADB\sim △ABC$ $\left[Henceproved\right]$

(iii) In $△BDC$ and $△ABC$, we have:

$\angle BDC=\angle ABC$ [Each equal to ${90}^0$]

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

So, by AA-similarity criterion, $△BDC\sim △ABC$ $\left[Henceproved\right]$

(iv) From (i), we have

$△ADB\sim △BDC$

$\therefore$ $\frac{AD}{BD}=\frac{BD}{DC}$

$\Rightarrow$ ${BD}^2=AD\times DC$ $\left[Henceproved\right]$