Question

. In figure angle ACB = 90 , CD perpendicular to AB, prove that CD2 = BD . CDC
A D B

Answer 1

Solution :

Consider ∆ACD and ∆ACB.

Angle CAD = Angle CAB ,

Angle CDA = Angle ACB ,

Hence , ∆ACD ~ ∆ABC (by AA similarity criterion ).

$= > \frac{AC}{AB} = \frac{AD}{AC}$

$=> AC^{2} = AB \times AD...............(i)$

Similarly , ∆BCD ~ ∆BAC ( by AA similarity criterion ).

$=> \frac{BC}{BA} = \frac{BD}{BC}$

$=> BC^{2} = BA \times BD...............(ii)$

On diving 2nd with the 1st , we get :-

$=> \frac{BC^{2} }{AC^{2} } = \frac{AB \times BD}{AB \times AD}$

$=> \frac{BC^{2} }{AC^{2} } = \frac{BD}{AD}$

Hence , Proved!

Answer 2

$\bold \blue {{GIVEN;}}$

ABC is a right angled triangle, in which /_C = 90° and CD_|_AB or the midpoint of AB.

$\bold \blue {{TO\: PROVE;}}$

2DC = AB

$\bold \blue {{CONSTRUCTION;}}$

A line through the mid point D of the hypotenuse AB and parallel to BC intersect at M such that MD||BC.

$\bold \blue {{PROOF;}}$

By construction, MD||BC and CM is a transversal.
Therefore, /_AMD = /_ACB (corresponding angles)
But, /_ACB = 90°
Therefore, /_AMD = 90°
=>DM_|_AC

Now, in AMD and CDM, we have DM = MD (common side)
AD = CD (since M is the mid point of AC)
/_ AMD = /_DMC (each 90°)
Therefore, ADM ~=CDM (by SAS congruence rule)
Then, CM = AM (by cpct) ............(i)
Also, given D is the mid point of AB.
Therefore, AD=BD=1/2AB............(ii)

From equation. (i) and (ii), we get
CD = AD= 1/2AB

$\bold {{HENCE,\: PROVED.}}$

$\bold \green {{HOPE \: YOU \: GOT \: IT}}$