Question

class 10 maths 2 1.similarity​
give correct answer

Answer 1

Hello, Buddy!!

||Required Response||

Given,

∠ABC = ∠EDC → 75°

In ∆ABC & ∆EDC

∠ABC = ∠EDC [Given]

∠ACB = ∠ECD [Common Angle]

∆ACB ~ ∆ECD [By AA Similarity]

Now, ∠ABC = ∠EDC

∠CAB = ∠CED

∠ACB = ∠ECD

AB/ED = AC/EC = CB/CD [By CPST]

CPST - Corresponding Parts of Similar Triangles.

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Hope It Helps You ✌️

Answer 2

Solution :-

Given :

• ∠ ABC = 75°
• ∠ EDC = 75°

In ∆ABC and ∆EDC

$\rm{∠ABC \cong ∠EDC \: \: \: \: ... \{given \: and \: each \: 75 \degree \} }$

$\rm{∠C \cong ∠C \: \: \: \: \: \: \: \: \: \: \: \: ... \{common \: angle \}}$

$\small \therefore\rm{∠ABC \sim ∠EDC \: by \: AA \: test \: \{here \: ABC} \leftrightarrow EDC \}$

$\rm{ \frac{AB}{ED} = \frac{BC}{DC} = \frac{AC}{EC} } \: \: \: ... \{CCST \} \: and$

Hence,

$\rm\red{\angle BAC \cong \angle DEC \: \: \: \: \: \: \: ... \{ CAST\}}$

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