Question

In a triangle B is on obtuse angle. If ADI CB Prove that Ac^2 = AB^2+ BC^2 + 2.BC.BD​

Answer 1

$\huge \mathcal \red {\underline{\underline{A}}} \huge \mathcal \green {\underline{\underline{N}}} \huge \mathcal \pink {\underline{\underline{S}}} \huge \mathcal \blue {\underline{\underline{W}}} \huge \mathcal \orange {\underline{\underline {E}}} \huge \mathcal \purple {\underline{\underline{R}}} ✧$

Given:-

A obtuse triangle ️ ABC, obtuse angle at B, and AD is perpendicular to BC.

To Prove :-

AC² = AB² + BC² + 2.BC.BD

Solution :-

In right triangle ️ ADB

Using Pythagoras Theorem

AB² = AD² + BD² ..........(1)

In right triangle ️ ADC

Using Pythagoras Theorem

AC² = AD² + CD²

AC² = AD² + (BD + BC)²

AC² = AD² + BD² + CB² + 2×BD×BC

AC² = AB² + BC² + 2×BD×BC [USING (1)]

Hence, Proved

$\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}$

Answer 2

Answer:

this is correct...................