Question

Triangle ABC is a right triangle right angled at B such that angleBCA = twice of angleBAC. Show that AC = twice of BC ​

Answer 1

Step-by-step explanation:

$<svg width="300" height="300" viewBox="0 0 100 100">\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ <path fill="pink" d="M92.71,7.27L92.71,7.27c-9.71-9.69-25.46-9.69-35.18,0L50,14.79l-7.54-7.52C32.75-2.42,17-2.42,7.29,7.27v0 c-9.71,9.69-9.71,25.41,0,35.1L50,85l42.71-42.63C102.43,32.68,102.43,16.96,92.71,7.27z"></path>\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ <animateTransform \ \textless \ br /\ \textgreater \ attributeName="transform" \ \textless \ br /\ \textgreater \ type="scale" \ \textless \ br /\ \textgreater \ values="1; 1.5; 1.25; 1.5;1.75;1.75 ;1.25; 1;" \ \textless \ br /\ \textgreater \ dur="2s" \ \textless \ br /\ \textgreater \ repeatCount="20"> \ \textless \ br /\ \textgreater \ </animateTransform>\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ </svg>$

Hey there,

Using angle sum property of triangle ABC

∠ABC + ∠BAC + ∠ACB = 180°

⇒ 90° + ∠BAC + 2∠BAC = 180°

⇒ 3∠BAC = 90°

⇒∠BAC = 30°

∴ ∠ACB = 2 × 30° = 60°

Similarly, it can be proved that ∠ADB = 60° and ∠BAD = 30°

∴ ∠CAD = 30° + 30° = 60°

In ∆ACD, ∠CAD = ∠ACD = ∠ADC = 60°

So, ∆ABC is an equilateral triangle.

∴ AC = CD = AD

It is know that the perpendicular drawn to a side from opposite vertex bisects the side

∴ CD = 2BC

⇒ AC = 2BC [CD = AC]

Hence, proved.

===========================

Hope this helped u..

PLZ MARK AS BRAINLIEST..

==================================================================================

Hope this will help you...

✔️❤__________________❤✔️

Answer 2

Answer:

the upper user is deceiver don't mark it brainliest