Question

In the given figure if a perpendicular is drawn from the right-angle vertex of the 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 length of the two parts of hypotenuse.

Answer 1

Hint: To solve the question, we have to apply the AAA symmetry, triangle properties for the given triangles to arrive at the conclusions. To solve the other statement, apply the properties of similar triangles to arrive at the proof.

Complete step-by-step answer:

Consider triangles \[\Delta ADB,\Delta ABC\]
\[\angle A=\angle A\]
Since A is the common angle in the both triangles.
\[\angle ADB=\angle ABC={{90}^{0}}\]
We know that the sum of angles of a triangle is equal to \[{{180}^{0}}\]. Thus, we get
\[\begin{align}
  & \angle A+\angle ABD+\angle ADB={{180}^{0}} \\
 & \angle A+\angle ACB+\angle ABC={{180}^{0}} \\
\end{align}\]
Thus, by solving these equations and by applying the above obtained we get
\[\begin{align}
  & \angle A+\angle ABD+{{90}^{0}}={{180}^{0}} \\
 & \angle A+\angle ACB+{{90}^{0}}={{180}^{0}} \\
 & \Rightarrow \angle A+\angle ABD+{{90}^{0}}=\angle A+\angle ACB+{{90}^{0}} \\
 & \therefore \angle ABD=\angle ACB \\
\end{align}\]
We know that AAA (Angle- Angle-Angle) symmetry that two triangles are similar, when all the corresponding angles of the triangles are equal.
Thus, we get \[\Delta ADB,\Delta ABC\] are similar triangles.
\[\Delta ADB\sim \Delta ABC\] ….. (1)

Consider triangles \[\Delta BCD,\Delta ABC\]
\[\angle C=\angle C\]
Since C is the common angle in the both triangles.
\[\angle BDC=\angle ABC={{90}^{0}}\]
We know that the sum of angles of a triangle is equal to \[{{180}^{0}}\]. Thus, we get
\[\begin{align}
  & \angle C+\angle CBD+\angle BDC={{180}^{0}} \\
 & \angle C+\angle CAB+\angle ABC={{180}^{0}} \\
\end{align}\]
Thus, by solving these equations and by applying the above obtained we get
\[\begin{align}
  & \angle C+\angle CBD+{{90}^{0}}={{180}^{0}} \\
 & \angle C+\angle CAB+{{90}^{0}}={{180}^{0}} \\
 & \Rightarrow \angle C+\angle CBD+{{90}^{0}}=\angle C+\angle CAB+{{90}^{0}} \\
 & \therefore \angle CBD=\angle CAB \\
\end{align}\]
We know that AAA (Angle- Angle-Angle) symmetry that two triangles are similar, when all the corresponding angles of the triangles are equal.
Thus, we get \[\Delta BDC,\Delta ABC\] are similar triangles.
\[\Delta BDC\sim \Delta ABC\] ….. (2)
We know that if one triangle is similar to two other triangles, then the other two triangles are similar to each other.
Thus, from equation (1) and equation (2), we get
\[\Delta ADB\sim \Delta BDC\]
Thus, the triangle on each side of the perpendicular is similar to each other and to the original triangle.
We know that in similar triangles, the ratio of corresponding sides is equal. Thus, we get
Consider triangles \[\Delta ADB,\Delta ABC\]
\[\dfrac{AD}{BD}=\dfrac{AB}{BC}\]
Consider triangles \[\Delta BDC,\Delta ABC\]
\[\dfrac{BD}{DC}=\dfrac{AB}{BC}\]
Thus, by solving the above equations, we get
\[\begin{align}
  & \dfrac{AD}{BD}=\dfrac{BD}{DC} \\
 & \Rightarrow AC.DC=BD.BD \\
 & \therefore AC.DC=B{{D}^{2}} \\
\end{align}\]
Thus, the square of the perpendicular is equal to the product of the length of the two parts of hypotenuse.

Note: The possibility of mistake can be, not applying the AAA symmetry, triangle properties for the given triangles to arrive at the conclusions. The other possibility of mistake can be not applying the properties of similar triangles to arrive at the proof of the statement, the square of the perpendicular is equal to the product of the length of the two parts of hypotenuse.