Question

In ,the figure AD perpendicular to BD and if AB square=A square+3BCsquare Prove that B
C=CD​

Answer 1

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Step-by-step explanation:

Given:

In ∆ABC , <C is an obtuse angle.

AD is perpendicular to BC.

and

AB² = AC²+3BC²

To prove:

BC = CD

Proof:

i) In ∆ADC , <D = 90°

AB² = AD² + DB² ---(1)

( By Phythagorean theorem )

ii) In ∆ADC , <D = 90°

AC² = AD² + DC² ----(2)

/* Subtract (2) from (1), we get

AB² - AC² = DB² - DC²

=> AC²+3BC²-AC²=(DC+BC)²-DC²

=> 3BC² = (DC+BC+DC)(DC+BC-DC)

/* By algebraic identity:

x²-y² = (x+y)(x-y) */

=> 3BC²=(2DC+BC)×BC

=> 3BC = 2DC+BC

=> 3BC - BC = 2DC

=> 2BC = 2CD

=> BC = CD

Hence proved.