Question

hey mate ,

Solve question 37 !! Class 10th ...
❌No spams please !!!❌​

Answer 1

Step-by-step explanation:

In a quadrilateral ABCD,

Given :- ∠B = 90°

AB² + BC² + CD² = AC² + CD² = AD²

{ ∴ In triangle ABC , AB² + BC² = AC² }

So, in triangle ACD, angle opposite to AD = 90° (By Converse of Pythagoras theorem)

⇒ ∠ACD = 90°

Hope it helps!

Answer 2

Answer:

Step-by-step explanation:

Sorry  bro/sis , took a long time as I have to install a graphing calculator, so that I could make an about accurate diagram for you. :)

Now, see the diagram and see this answer. I hope it will help you. :)

$AD^{2}=AB^{2}+BC^{2}+CD^{2}$

Now, to prove < $ACD$=90, we can simply prove that

DC^{2}+AC^{2}=AD^{2}

so, let's begin!

$AD^{2}=AB^{2}+BC^{2}+CD^{2}\\\\\\so, \\AD^{2}=( AB^{2}+BC^{2})+CD^{2}\\ => AD^{2}=AC^{2}+CD^{2} (AB^{2}+BC^{2}=AC^{2}\;By\;baudhyan's\;theorem.\\i.e. \;Pythagora\;theorem)\\\\ Since\;\;AD^{2}=AC^{2}+CD^{2} , But\;that\;is\;only\;possible\;when\;\\<ACD=90degree\\\\So,\;by\;converse\;of\;pythagoras\;theorem,\\<ACD=90degree$

Do mark as $BRAINLIEST$ if it helps you.

$LEARNING \;\;TOGETHER\;:)$

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