Question

If in Triangle ABC, AB > AC and AD perpendicular to BC, prove that AB^2 – AC^2 = BD^2 – CD^2.​

Answer 1

Answer:

AB^C-AC^2=ABC^3-CD^2=ABC5

Hope to clear your answer

Answer 2

Step-by-step explanation:

After drawing line AD perpendicular to BC

you have two right angle triangle

∆ADC and ∆ADB

Remember to take common sides AD and compare.

in ∆ADC,

AC²=AD²+DC²

AD²=AC²- DC² .........eq(1).

Now in ∆ADB

AB²=AD²+DB²

AD²=AB²-DB²...........eq(2)

from equation 1 and 2 we have,

AC²-DC²=AB²-DB²

AB²-AC²=BD²-DC²

Hence, proved.