Question

In fig, \[ABD\] is a triangle right angled at \[A\] and \[AC \bot BD\]. Show that,
1. \[A{B^2} = BC.BD\]
2. \[A{C^2} = BC.DC\]
3. \[A{D^2} = BD.DC\]
               

Answer 1

Hint: At first, using the given condition we will try to find that the triangles are similar.
If the three angles of any triangle are equal to the respective angles of another triangle, then the triangle is called a similar triangle.
Then, by the common side of a similar triangle we can prove the required proof.

Complete step-by-step answer:
It is given that; \[ABD\] is a triangle right angled at \[A\] and \[AC \bot BD\]
We have to show that,
1. \[A{B^2} = BC.BD\]
2. \[A{C^2} = BC.DC\]
3. \[A{D^2} = BD.DC\]
Since, \[AC \bot BD\], \[\Delta ABC\] and \[\Delta BAD\] are right-angle triangles.
So, \[\angle BCA = \angle DAB = {90^ \circ }\]
Again, \[\angle B\] is the common angle of \[\Delta ABC\] and \[\Delta BAD\].
Now, in \[\Delta ABC\] and \[\Delta BAD\], we have,
\[\angle BCA = \angle DAB = {90^ \circ }\]
\[\angle B\] is the common angle
So, by A-A-A condition, \[\Delta ABC\] and \[\Delta BAD\] are similar triangles.
From the common side of similar triangle, we have,
\[\dfrac{{BC}}{{AB}} = \dfrac{{AC}}{{AD}} = \dfrac{{AB}}{{BD}}\]… (1)
And, from the common angle of similar triangle, we have,
\[\angle BAC = \angle BDA\]… (2)
From (1) we get,
\[\dfrac{{BC}}{{AB}} = \dfrac{{AB}}{{BD}}\]
By cross multiplication we get,
\[A{B^2} = BC.BD\]
Again, in \[\Delta ABC\] and \[\Delta CAD\], we have,
\[\angle BCA = \angle DCA\] (each angle is \[{90^ \circ }\])
From (2) we get, \[\angle BAC = \angle BDA\]
So, by A-A-A condition, \[\Delta ABC\] and \[\Delta CAD\] are similar triangles.
From the common side of similar triangle, we have,
\[\dfrac{{BC}}{{AC}} = \dfrac{{AC}}{{CD}} = \dfrac{{AB}}{{AD}}\]
So, \[\dfrac{{BC}}{{AC}} = \dfrac{{AC}}{{CD}}\]
By cross multiplication we get,
\[A{C^2} = BC.DC\]
Since, \[\Delta ABC\] and \[\Delta CAD\] are similar triangles and \[\Delta ABC\] and \[\Delta BAD\] are similar triangles.
So, \[\Delta CAD\] and \[\Delta BAD\] are similar triangles.
From the common side of similar triangle, we have,
\[\dfrac{{AB}}{{AC}} = \dfrac{{AD}}{{CD}} = \dfrac{{BD}}{{AD}}\]
So, \[\dfrac{{AD}}{{CD}} = \dfrac{{BD}}{{AD}}\]
By cross multiplication we get,
\[A{D^2} = BD.DC\]
Hence,
1. \[A{B^2} = BC.BD\]
2. \[A{C^2} = BC.DC\]
3. \[A{D^2} = BD.DC\]

Note: We know that, if the three angles of any triangle are equal to the respective angles of another triangle, then the triangles are called similar triangles. This is known as A-A-A condition.
If all the sides of any triangle are equal to the respective sides of another, then the triangles are called similar triangles. So, the ratio of the respective sides are equal.