Question

In the right triangle, B is a point on AC such that AB+AD=BC+CD.If AB=x,BC=h and CD=d then find x(in terms of h and d)

Answer 1

Answer:

hd/2h+d

Step-by-step explanation:

AB+AD=BC+CD

x+AD=h+d

AD=h+d-x

AD^2= AC^2+CD^2

AD^2=(x+h)^2 + d^2

AD=✓(x+h)^2+d^2

Remaining you can see in attached image

Answer 2

The value of $x$ is $\frac{hd}{2h+d}$

Step-by-step explanation:

Given,

In right triangle, $B$ is a point on $AC$ such that $AB+AD=BC+CD$ and $AB=x,BC=h and CD=d$

∵ $AB+AD=BC+CD$

Plug the value,

⇒$x+AD=h+d$

⇒$AD=h+d-x$ __1

Also, $\triangle ACD$ is a right angle triangle.

So,

$AD^{2} =AC^{2} +CD^{2}$

⇒$AD^{2} =(x+h)^{2} +d^{2}$  (∵$AC=AB+BC=x+h$ )

⇒$AD=\sqrt{(x+h)^{2}+d^{2}}$__2

From equation-1 and 2,

$h+d-x=\sqrt{(x+h)^{2}+d^{2}}$

⇒$(h+d-x)^{2}=(x+h)^{2}+d^{2}$

⇒$h^{2}+d^{2}+x^{2}+2hd-2dx-2hx=x^{2}+h^{2}+2hx+d^{2}$

⇒$2(hx+hx+dx)=2hd$

⇒$(2hx+dx)=hd$

⇒$x(2h+d)=hd$

∴ $x=\frac{hd}{2h+d}$

So, The value of $x$ is $\frac{hd}{2h+d}$