Question

In fig. \[D\] is a point on the hypotenuse \[AC\] of \[\Delta ABC\], such that \[BD \bot AC,DM \bot BC\] and \[DN \bot AB\]. Prove that:
1. \[D{M^2} = DN.MC\]
2. \[D{N^2} = DM.AN\]
               

Answer 1

Hint: Using the given condition we will try to find that the triangles are similar.
By the similarity rules, we can prove the required proof.
If the three angles of any triangle are equal to the respective angles of another triangle, then the triangle is called a similar triangle.

Complete step-by-step answer:
It is given that; \[D\] is a point on hypotenuse \[AC\] of \[\Delta ABC\], such that \[BD \bot AC, DM \bot BC\] and \[DN \bot AB\].
We have to proof that;
1. \[D{M^2} = DN.MC\]
2. \[D{N^2} = DM.AN\]
In \[\Delta ABC\], we have, \[DN \bot AB\] and \[BC \bot AB\]
So, we have, \[DN || BC\]… (1)
Again, \[DM \bot BC\] and \[BC \bot AB\]
So, we have, \[DM || AB\]… (2)
Therefore, from (1) and (2) we get,
\[DMBN\]is a rectangle.
So, we have, \[BM = DN\]
We know that the sum of all the angles of a triangle is \[{180^ \circ }\].
From, \[\Delta BMD\], we get,
\[\angle BMD + \angle BDM + \angle DBM = {180^ \circ }\]
Since, \[\angle BMD = {90^ \circ }\]
So, we get, \[\angle BDM + \angle DBM = {90^ \circ }\]…. (1)
Similarly,
From, \[\Delta CMD\], we get,
\[\angle CMD + \angle CDM + \angle DCM = {180^ \circ }\]
Since, \[\angle CMD = {90^ \circ }\]
So, we get, \[\angle CDM + \angle DCM = {90^ \circ }\]…. (2)
Again, it is given, \[BD \bot AC\]
So, \[\angle BDM + \angle MDC = {90^ \circ }\]… (3)
From (1) and (3) we have,
\[\angle BDM + \angle DBM = \angle BDM + \angle MDC\]
Eliminating the same angle from both the side we get,
\[\angle DBM = \angle MCD\]… (4)
Similarly, \[\angle BDM = \angle MCD\]… (5)
Now, from \[\Delta BMD\] and \[\Delta CMD\] we get,
\[\angle BMD = \angle DMC\] (as each angle is \[{90^ \circ }\])
\[\angle BDM = \angle MCD\] (from (4))
\[\angle DBM = \angle MCD\] (from (5))
By the A-A-A rule we get,
\[\Delta BMD \cong \Delta CMD\]
From the common side of similar triangle, we have,
\[\dfrac{{BM}}{{DM}} = \dfrac{{MD}}{{MC}}\]… (6)
\[\dfrac{{DN}}{{DM}} = \dfrac{{MD}}{{MC}}\]… (7)
From, (6) and (7) we get,
\[BM = DN\]
From (6) we get,
\[D{M^2} = DN.MC\]
Similarly, \[\Delta DNB \cong \Delta AND\]
\[\dfrac{{BN}}{{DN}} = \dfrac{{ND}}{{AN}}\]
Since, \[BN = DM\]
We get,
\[D{N^2} = DM.AN\]
Hence,
1. \[D{M^2} = DN.MC\]
2. \[D{N^2} = DM.AN\]
Here’s the proof.

Note: Angle sum property states that the sum of all the angles of any triangle is \[{180^ \circ }\].
If the three angles of any triangle are equal to the respective angles of another triangle, then the triangle is called a similar triangle.
The opposite sides of the rectangle are equal to each other.
If any two angles of a triangle are equal to any two angles of another triangle, then the two triangles are similar to each other