Question

solve corresponding part of congruent triangles​

Answer 1

Step-by-step explanation:

$\blue{\mathfrak{\underline{\large{Given}}}:} \\ ΔABD≈ΔCDB \\ m\angle DAB=65° \\ m\angle BDA=35° \\ AD=10cm \\ \blue{\mathfrak{\underline{\large{To \: find}}}:} \\ m\angle \: B \\ AC \: and \: BC \\ \blue{\mathfrak{\underline{\large{Finfing}}}:} \\ \red{\mathfrak{\large{since, \:we \: can \: see \: AD||BC \: and \: AB||DC }}} \\\blue{\mathfrak{\large{Therefore, \: ABCD \: is \: quadrilateral}}} \\ \angle BDA = \angle CBD = 35° \\ In \: ΔABD : \\ \angle DAB+\angle BDA+\angle ABD=180° \\ 65°+35°+\angle ABD=180° \\ \angle ABD=80° \\ \angle B = \angle ABD + \angle CBD \\ \angle B = 80° + 35° \\ m\angle B = 115° \\ \red{\mathfrak{\large{since, \:we \: can \: see \: diagonal \: BD \: bisect \: AC }}} \\ \blue{\mathfrak{\large{Therefore, \: ABCD \: is \: a \: paralleleogram}}} \\ =⟩ \: AD=BC=10cm \\ =⟩AB=DC \\ =⟩\angle B=\angle D \: and \: \angle A=\angle C \\ In \: ΔABC \: and \: ΔADC$

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