Question

prove that triangle ABC is corresponding to triangle DCB if AC = DB and angle A = angle D = 90 degree​

Answer 1

Answer:

In△ABCand△DCB

\begin{gathered} \angle A = \angle D = 90 \: degree ( given )\\\blue {( Right \:angle ) } \end{gathered}

∠A=∠D=90degree(given)

(Rightangle)

BC = CB \: \blue{( Hypotenuse )}BC=CB(Hypotenuse)

AC = DB \: ( Side )AC=DB(Side)

\triangle ABC \cong \triangle DBC△ABC≅△DBC

\pink { ( RHS \: Congruence )}(RHSCongruence)

\underline{ \blue { RHS \: congruence \:rule:}}

RHScongruencerule:

If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and corresponding side of the other triangle , then the two triangles are cong

Answer 2

Hence proved∆ABC~∆DCB