Question

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Answer 1

Your Answer:

$\tt In \ \ \triangle BDE$

BD = DE

So,

$\tt \angle{DBE} = \angle{DEB} ( \because equal \ \ sides \ \ subtends \ \ equal \ \ angles)$

So,

Using Angle sum property in triangle BDE

$\tt \angle BDE + \angle DBE + \angle DEB = 180^o \\\\ \tt \Rightarrow 2\angle DEB + 110^o = 180^o \\\\ \tt \Rightarrow 2\angle DEB = 70^o \\\\ \tt \Rightarrow \angle DEB = \dfrac{70}{2} \\\\ \tt \Rightarrow \angle DEB = 35^o$

$\tt In \ \ \triangle{D E F}$

DE = EF

So,

$\tt \angle EFD = \angle EDF \ \ (\because equal \ \ sides \ \ subtends \ \ equal \ \ angles )$

And

$\tt \angle EDF = 180 - 110 (\because linear \ \ pair) \\\\ \tt \Rightarrow \angle EDF = 70^o = \angle EFD$

Using Angle sum property in triangle DEF

$\tt 70^o+70^o + \angle {D E F} = 180^o \\\\ \tt \Rightarrow 140^o +\angle {D E F} = 180^o \\\\ \tt \Rightarrow \angle D E F = 40^o$

We have also

$\tt \angle BEF = \angle D E F + \angle BED \\\\ \tt \Rightarrow \angle BEF = 40^o + 35^o \\\\ \tt \Rightarrow \angle BEF = 75^o$

And

$\tt \angle ABE = \angle BEF ( \because alt. int angles) \\\\ \tt \Rightarrow\angle ABE = 75^o$

So,

$\tt \star \angle BED = 35^o\\\\ \tt \star \angle D E F = 40^o \\\\ \tt \star \angle ABE = 75^o$

Answer 2

Answer:

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