Question

..........................answer, please

Answer 1

Step-by-step explanation:

Given:- X is a point inside the square ABCD and AXYZ is also a square

To prove:- ΔABX ≅ ΔADZ

Proof:-

In square ABCD,

AB = AD ----- 1

also,

∠BAD = 90° ------ 2

In square AXYZ,

AX = AZ ------ 3

also,

∠XAZ = 90° ------ 4

Now,

Let ∠DAX = θ ----- 5

then ∠BAD = ∠BAX + ∠DAX

So, from eq.2 and eq.5

90° = ∠BAX + θ

∠BAX = 90° - θ ------ 6

Similarly,

∠XAZ = ∠DAX + ∠DAZ

from eq.4 and eq.5

90° = θ + ∠DAZ

So, ∠DAZ = 90° - θ ----- 7

Now, if we compare eq.6 and eq.7 we get

∠DAZ = ∠BAX ------- 8

Now, from ΔABCD and ΔAXYZ

AB = AD (from eq.1) (Side)

∠DAZ = ∠BAX (from eq.8) (Angle)

AX = AZ (from eq.3) (Side)

Thus,

ΔABX ≅ ΔADZ (By S.A.S Congruency)

Hope it helped and you understood it........All the best