Question

in an acute angle triangle ABC seg AD is perpendicular to side BC and B-D-C then prove that AB^2-BD^2=AC^2-CD^2​

Answer 1

Step-by-step explanation:

In a Triangle ABC with AD perpendicular to BC, two right angled Triangles are formed Viz. Triangle ADB and Triangle ADC.

Since these are Right Angled triangles, we have

AB^2 = BD^2 + AD^2 —— 1 and

AC^2 = CD^2 + AD^2 —— 2

From 1, we get

AD^2 = AB^2 - BD^2 —— 3

and from 2, we get

AD^2 = AC^2 - CD^2 —— 4

Since both 3 and 4 are about AD^2, we get

AB^2 - BD^2 = AC^2 - CD^2

On rearranging BD^2 and CD^2, we get

AB^2 + CD^2 = AC^2 + BD^2

Hence proved!