Question

please help tomorrow my exam

Answer 1

Answer:

Step-by-step explanation:

Draw AE perpendicular to BC.

Since angle AED = 90°,

Therefore, in triangle ADE

angle ADE < 90° and angle ADB > 90°.

Thus, triangle ABD is obtuse angled triangle and triangle ADC is acute angled triangle.

Triangle ABD is obtuse angled at D and AE is perpendicular to BD produced,

Therefore,

AB2 = AD2 + BD2 + 2 BD x DE (i)

Triangle ACD is acute angled at D and AE is perpendicular to BD produced,

Therefore,

AC2 = AD2 + DC2 - 2 DC x DE

AC2 = AD2 + BD2 - 2 BD x DE (ii) (since CD = BD)

Adding (i) and (ii)

AB2 + AC2 = 2AD2 + 2BD2

AB2 +AC2 = 2(AD2 + BD2)

Hope it helps you for your exam....