Question

Guys please help me solving this 10th one I am not getting it..

Answer 1

Question given :

In the adjoining figure prove that ∆ABC is congruent to ∆DCB , if AC = DB and angle A = angle D = 90°

To prove :

Triangle ∆ABC is congruent to ∆DCB

Required Solution :

AC = DB = Given

Angle D = Angle A = 90°

BC = BC = Common

Thus , ∆ABC is congruent to ∆DCB by RHS congruency rule

Required knowledge approach :

• SSS congruency rule - Side side side
• RHS congruency rule - Right angle Hypotenuse Side
• ASA congruency rule - Angle side angle
• SAS congruency rule - Side Angle Side

Answer 2

$\huge{\pink{\underline{\underline{\bf{\bigstar\;SoluTion:}}}}}$

Given,

The Triangles ABC & DBC are standing on the same base of BC, AC = DB & Angle A = Angle D.

In ∆ABC & ∆DBC

$→\;{\bf{AC = DB}}$ (Given)

$→\;{\bf{Angle\;A = Angle\;D\;☞\;90°}}$ (Given)

$→\;{\bf{BC = BC}}$ (Common Side)

☞ Hence, ∆ABC is Congruent to ∆DBC by RHS Congruence Rule.

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Extra Information ℹ️

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