Please help me solving this question in easiest way you can!!
Attachments:
Answers
Answered by
1
Hi✌️✌️
Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB2 = AD2 + BD2 ...(i)
AC2 = AD2 + DC2 ...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2 - AC2 = BD2 - DC2
= 9CD2 - CD2 [∴ BD = 3CD]
= 9CD2 = 8(BC/4)2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB2 - AC2 = BC2/2
⇒ 2(AB2 - AC2) = BC2
⇒ 2AB2 - 2AC2 = BC2
∴ 2AB2 = 2AC2 + BC2.
Please mark me as brainliest
Similar questions