Question

answer it please may god help u​

Answer 1

$\huge \fbox \red{Answer}$

i) In ∆BAC and ∆DAC

AC = AC (common)

∠BCA = ∠DCA (90°)

DC = BC (AZ bisects it)

∴ ∆BAC ≈ ∆DAC (SAS test)

ii) ∆BAC ≈ ∆DAC (Proved above)

iii) ∆BAC ≈ ∆DAC [proved in (i)]

AB = AD (cpct)

Answer 2

Answer :-

i) Three pair of equal parts in triangle BAC and DAC -

→ ∠ DAC = ∠ BAC

→ ∠ DCA = ∠ BCA

→ AC = AC

ii)

→ ∠ DAC = ∠ BAC

→ AC = AC

→ ∠ DCA = ∠ BCA

By SAS congruence criterion :-

∆ BAC ≅ ∆ DAC

iii)

∆ BAC ≅ ∆ DAC

So, By CPCT :-

→ AB = AD