Question

in a triangle pqr PD is perpendicular on QR such that D lies on QR. If PQ=a,PR=b,QD=c and DR=d,prove that (a+b)(a-b)=(c+d)(c-d)

Answer 1

hope this will help you.....

Answer 2

It shows that (a + b)(a - b) = (c + d)(c - d) is proven in triangle PQR in which PD is perpendicular to QR.

Solution:

As shown in triangle PQR, PD is perpendicular on QR

Now as given in question the value of  

PQ = a

PR = b

QD = c  

DR = d

To find value of PD:  

$\begin{array}{l}{P D=P Q^{2}-Q D^{2}} \\ {P D=P R^{2}-D R^{2}}\end{array}$

Now putting the value in $P D=P Q^{2}-Q D^{2}$

$P D=P R^{2}-D R^{2}$

We get

$\begin{array}{l}{P D=a^{2}-c^{2}} \\ {P D=b^{2}-d^{2}}\end{array}$

Now equating we get

$\begin{array}{l}{a^{2}-c^{2}=b^{2}-d^{2}} \\ {a^{2}-b^{2}=c^{2}-d^{2}}\end{array}$

(a + b) (a - b) =(c + d) (c - d)  

Hence proved.