Question

in the given figure X is any point within a square ABCD on AX square AXYZ is described prove that the BX = DZ

Answer 1

Given :

AXYZ is a square

ABCD is a square

To Prove :

BX = DZ

Proof :

In $\triangle{BAX} \: and \: \triangle{ZAD}$

(S) ZA = AX ______{side of square AXYZ}

(A) $\angle{BAX} \:=\: \angle{ZAD}$ by eq 2

(S) AD = AB ______{side of square ABCD}

Now we will consider,

Let $\angle{XAD} \: =\: x$

$\angle{ZAD} \: +\: x\:=\: 90$

$\angle{ZAD} \:=\: 90\: - \: x$ - - - from(eq1)

Other one

$\angle{BAX} \:+\: x\:=\: 90$

$\angle{BAX} \:=\: 90\: - \: x$ put values

$\angle{BAX} \:=\: \angle{ZAD}$ by eq 1 and this is eq 2

Congruence rule is SAS

$\triangle{BAX} \: \cong \: \triangle{ZAD}$

$\boxed{BX = DZ}$ by Cpct (Corresponding parts of Congruent Triangle)

Hence Proved

Answer 2

Answer:

helo

Step-by-step explanation:

can u pls tellfrom where this question is taken,as u have already got th answer?

ill mark you brainliest.pl pls ls tell yr thank you