Question

Please ans the question urgent......
17 and 19

Answer 1

$\boxed{QUESTION:-}$

17)↑

$\boxed{SOLUTION:-}$

$\mathbb{Given:-}$

<ACB=<CDA

AC=6CM

AD=3CM

$\blue{PROOF:-}$

In ∆ACB & ∆ACD

=> <A=<A[common]

=> <ACB=<CDA[Given]

=>∆CAB≈∆CDA [SIMILAR]

=>$\frac{AC}{AD}$=$\frac{AB}{AD}$

PUTTING THE VALUES,

$\frac{6}{3}$=$\frac{AB}{3}$

2=$\frac{AB}{3}$

Cross Multiply↓

=>AB=6cm

$\mathcal{BE \: BRAINLY}$