Question

Observe the figure. In A ABC, AD is
altitude and BD is half of CD, then
show that AC2 = 3BD2 + AB.​

Answer 1

LET

BD=X

DC=2X

2x sq=AC sq-AD  sq----------------1

X sq=AB SQ+AD SQ---------------2

SUBSTRACT 1_-_2 WE GET

2X SQ-X SQ=AC SQ-AB SQ

SIMPLIFY

WE GET

3X SQ=AC SQ-AB SQ

RE ARRANGE AND PUT X WE GET

AC2 = 3BD2 + AB2.​

Answer 2

proved

Step-by-step explanation:

Given : In A ABC, AD is

altitude and BD is half of CD

to prove : AC² = 3BD² +AB

proof

let BD = x

and given that BD is half of CD therefore

CD = 2x

now , in triangle ADC

AC² = CD²+AD²

AC²= (2x)²+AD²

AC²= 4x²+AD²..(1)

now , in triangle ABD

AB² = AD²+BD²

AB² = AD²+ x²....(2)

subtract 2 from 1 we get

AC² -AB² = 4x²+AD² - (AD²+ x²)

AC² -AB² =  3x²

x = BD

therefore

AC² -AB² =   3BD²

AC² = 3BD² +AB

hence , proved

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